Quantity of heat (Calorimetry, Change of state) | Conceptual notes | Solutions to short questions and Numerical problems | Grade 11 (Physics) | Physics in Depth
- Calorimetry is the experimental technique for quantitative measurement of heat exchange.
-
1 cal = 4.2 Joule - The amount of heat absorbed depends on the mass of the body, the change in temperature, the material of the body as well as the surrounding conditions such as pressure, etc.
\[Q=ms \Delta \theta\] where \(\Delta \theta\) is the change in temperature, m is the mass of the body, Q is the heat supplied, and s is the specific heat capacity. - Specific heat capacity , s , of a substance is defined as the amount of heat required to change the temperature of unit mass of substance through one degree. It's unit is Jkg\(^{-1}\) K\(^{-1}\) in SI systems and cal g\(^{-1}\) \(^\circ\) C\(^{-1}\).
- Heat capacity or thermal capacity of a substance is defined as the amount of heat required to change its temperature through one degree. Mathematically,
\[Q=ms\] It's unit is J K\(^{-1}\) in SI units and cal \(^\circ\) C\(^{-1}\) in CGS units. - Water equivalent of a substance is the mass of water that will absorb or lose the same quantity of heat as the substance for the same change in temperature.
If w is the water equivalent of a substance, then ,
\[w=\frac{ms}{s_w}\] Here, m is the mass of the substance, s is its specific heat capacity and \(s_w\) is the specific heat capacity of water. In CGS system, \(s_w\) = 1 cal g\(^{-1}\) \(^\circ\) C\(^{-1}\).
Thus,
\[w=ms\]
So, water equivalent of a substance is numerically equal to its thermal capacity in CGS system. But the unit of water equivalent is grams and that of thermal capacity of calories. - Neglecting any heat exchange with the surrounding, the principle of calorimetry states that the total amount of heat lost by the hot substance is equal to the heat gained by
the cold substance.
i.e.,heat loss = heat gain
- Newton's law of cooling states that the rate of loss of heat of a liquid is directly proportional to the difference in temperature between the liquid and the suroundings.
\[\frac{dQ}{dt} \propto (\theta-\theta_0)\] where dQ is the loss of heat in small time dt and \(\frac{dQ}{dt}\) is the heat lost. - Latent heat of a substance is the amount of heat required to change the state of unit mass of the substance from solid to liquid or from liquid to vapour without any change in temperature. It is the heat which is hidden from the thermometer. Latent heat is represented by L and its unit is J kg\(^{-1}\) or cal g\(^{-1}\).
- The latent heat of fusion of ice is defined as the amount of heat required to change unit mass of ice from 0\(^\circ\) C to water at same temperature. It's value is 3.36 \(\times\) 10 \(^5\) J kg\(^{-1}\) in SI units or 80 cal g\(^{-1}\) in CGS system.
- The latent heat of vaporization of a liquid is defined as the amount of heat required to change unit mass of liquid at boiling point into vapour at the same temperature. It's value is 2.26 \(\times\) 10\(^6\) J kg\(^{-1}\) in SI units or 540 cal g\(^{-1}\) in CGS system.
- Supercooling is the process of cooling liquid below its freezing point without solidification or crystallization.
- Experimentally, it has been found that in case of substances which expand on solidification, the increase of pressure lowers their melting point and in case of substances which contract on solidification, the increase of pressure increases their melting point .
- Regelation is the phenomenon of melting of ice due to compression and freezing after removal of applied pressure.
Short Answer Questions Solution
- Why do we feel cool in the mouth when we eat halls?
Solution:
When we eat halls, the ingredients of halls vaporize by taking the necessary amount of heat from our mouth. This heat is called the latent heat of vaporization that is equal to 540 cal/g. Thus, due to the loss of heat, we feel cool in our mouth while eating halls. - Water remains cool in earthern pots than in metal pots in summer. Explain why?
Solution:
Earthern pots has many holes in it through which the water seeps out and evaporates. The energy needed for evaporation is taken from the water kept in the earthern pot (i.e., Latent heat of vaporization). Due to this reason, water remains cool in earthern pots than in metal pots in summer. - During winter, the animals curl into a ball. Explain, why?
Solution:
The total energy radiated by the body depends on its surface area. More the surface area, more is the heat radiated from the body. So, during winter when animals feel very cold, they curl their body to decrease the surface area and so the heat radiated. - How can water be boiled in a paper cup?
Solution:
When we heat the water in a paper cup, the water absorbs the heat and doesn't let the paper cup reach its ignition temperature. In this process, the water boils even without letting the paper cup burn. - Why does food cook faster in a pressure cooker than in an open pot?
Solution:
With the increase in pressure, the boiling point of water in the pressure cooker increases above 100\(^\circ\) C. So, the heat contained by the mass of water increases and thus the food is cooked in very less time in a pressure cooker. - If you add heat to an object , do you necessarily increase its temperature? Justify your answer.
Solution:
It is not necessary that the temperature is increased as the heat is added to the object. For example, when water boils at 100\(^\circ\) C, adding heat to it changes its state to vapor without increase in temperature i.e., at this point, the heat added to the water is used in change of state. - Explain the significance of anomalous expansion of water with an example observed in nature.
Solution:
Most liquids expands on heating and contracts on cooling. Though, water has a different behaviour. As we lower the temperature of water from room temperature, first the volume of water decreases upto 4\(^\circ\) C but below this temperature, it's volume goes on increasing and becomes maximum at 0\(^\circ\) C. This peculiar nature of water is the anomalous expansion of water.
Significance:
4\(^\circ\) C water has minimum volume and maximum density whereas 0\(^\circ\) C water has maximum volume and minimum density. So, in cold countries, the water surfaces like pond remains cold with the presence of ice but the lowest layer of water is warm. Because the temperature of the lower layer never falls below 4\(^\circ\) C, even though the upper layer freezes into ice. Since ice is a bad conductor of heat, it provides the heat shield and does not allow the heat to escape out. Due to this reason, the aquatic plant and animal life can stay alive in frozen pond. - Groundnuts are fried along with sand, why?
Solution:
The specific heat capacity of sand is very low (i.e., 840 J kg\(^{-1}\) \(^\circ\) C\(^{-1}\)) and thus it heats very quickly (absorbing less amount of heat). Due to this reason, the groundnuts are fried fast when fried with sand. - During a high fever, a wet cloth is kept on the forehead of a person. Why?
Solution:
Water has very high specific heat capacity i.e., it absorbs or lose large amount of heat before heating or cooling. When a wet cloth is kept in forehead of a person with the fever, it absorbs large amount of heat from the person's forehead and temperature of the high fever person falls down quickly. This gives relief to the person as he looses large amount of heat. Thus, a wet cloth is kept on the forehead of a person during high fever. - Why is a spark, produced when two stones are stricken against each other?
Solution:
When two stones are stricken against each other, the friction between the rocks produces heat. This heat appears as a spark. - The length of a brass rod is found to be smaller on a hot summer day than on a cold winter day as measured by the same aluminium scale.
Do we conclude that brass shrinks on heating?
Solution:
The linear expansion of brass rod and aluminium scale is not same. Eventhough both the rods expands on hot summer day, brass rod appears to be smaller when measured with aluminium scale because the aluminium expands more than brass rod because of its greater linear expansivity.i.e., \[\alpha_{brass}=19\times 10^{-6} / ^\circ \text{C}\] \[\alpha_{alum}=25\times 10^{-6} / ^\circ \text{C}\] So, it can't be concluded that the brass rod shrinks on heating. The linear expansivity of this two materials should be kept into account. - If an automobile engine is overheated, it is cooled by putting water on it. It is advised that the water should be put slowly with engine running. Explain the reason. (Hint! High specific heat capacity of water)
- A spherical shell is heated. The volume changes accordidng to the equation V\(_\theta\) = V\(_0\)(1+\(\gamma\theta\)). Does the volume refer to the
volume enclosed by the shell or the volume of the material making up the shell?
Solution:
Here, the volume refers to the material making up the shell. The volume of the material having cubical expansivity \(\gamma\) changes as the temperature is changed. - Why do we feel cold when we spray perfume on our body?
Solution:
When the perfume is sprayed on our body, it vaporizes very quickly by taking the necessary amount of heat from our body. So, as the body loses heat, we feel cold. - When you come out of swimming pool, you feel cold. Why? (Hint! Latent heat of vaporization)
- Why do we think that the latent heat of vaporization is so much larger than the latent heat of fusion of substance?
Solution:
The volume of the substance increases more when liquid changes to vapor state compared to the change from solid to liquid. So, the large amount of work has to be done against the surrounding pressure for vaporization. Due to this reason, the latent heat of vaporization is much larger than the latent heat of fusion of substance. - Why does steam at 100\(^\circ\) C cause severe burns than that water at 100\(^\circ\) C?
Solution:
When water changes its state from liquid to gas (i.e., steam), the additional amount of heat (i.e., latent heat of vaporization) is required. So, steam carries more heat than the water at the same temperature. Thus, steam at 100\(^\circ\) C cause severe burns than that water at 100\(^\circ\) C. - Dogs in summer keep their tongues out. Explain.
Solution:
When dogs keep their tongues out during summer, the water in its tongue vaporizes by taking the required latent heat from it. As dogs looses heat through its tongue during this process, dogs feel cold. So, dogs keep their tongues out during summer. - Why is ice cream colder than ice?
Solution:
Ice cream contains many impurities such as salt, so the melting point of ice cream is lower than the melting point of ice. Due to this reson, the ice cream melts below 0\(^\circ\) C. Thus, ice cream is colder than ice. - We feel cold during the melting of snow than during the snowfall, why?
Solution:
When snow melts, it uses up the heat from the surrounding to melt(i.e., latent heat of fusion). So, the surrounding looses heat (temperature falls down) and we feel cold while snow melts. - Can we walk on the surface of frozen lake?
Solution:
When we walk on the surface of frozen lake, the increase in pressure below our feet lowers the melting point (i.e., regelation phenomena). Due to it, a thin layer of ice melts forming water which makes the surface slippery so that it is difficult to walk. - Why is salt sprayed on the road covered with the snow in cold countries?
Solution:
Impurities such as salt lowers the melting point of ice and thus it melts below 0\(^\circ\) C. So, when the salt is sprayed in the snowy lake, snow melts below its melting point. Due to this reason, the salt is sprayed on the snowy lake. - Define regelation.
Solution:
The phenomenon of melting of ice due to compression and freezing after removal of applied pressure is called regelation. For example, loop a fine wire around a block of ice, with a heavy weight attached to it. At the portion where the wire passes, the pressure increases and so, the melting point of ice decreases. As the wire passes inside the ice block, the water above it freezes. So, the wire passes over the block over it without separating it. - Why is it harmful to dry wet clothes on our body?
Solution:
While drying wet clothes on our body, the water evaporates off the cloth by taking the required latent heat of vaporization from our body and thus our body looses heat. Due to this reason, it is not advised to dry wet clothes on our body. - Why is a baby wrapped tightly with warm clothes than an adult man in winter?
Solution:
From the Newton's law of cooling, \[log(\theta-\theta_0)=\frac{-k}{ms}t+C\] Here, \(\theta\) is the temperature of the body, \(\theta_0\) is the temperature of the surrounding, m is the mass of body and s is the specific heat capacity.
k is proportional to the surface area of the body and m is proportional to the volume of body and thus the ration \(\frac{k}{ms}\) is inversely proportional to the linear dimension. Due to this reason, the rate of heat lost from the body with less linear dimension is very rapid (compare with \(y=mx+c\), m is the slope and in our case \(\frac{-k}{m}\) is the slope and with the decrease of the ratio \(\frac{k}{m}\), there is increase in rate of heat loss). Thus, a baby is wrapped tightly with warm clothes than an adult man in winter. - Why do animal eat more during winter?
Solution:
The temperature difference between the surrounding and the body of animal is very large in winter compared to the summer. From Newton's law of cooling, the rate of heat loss by a body is directly proportional to the temperature difference between the surrounding and the body. So, animal looses more heat during winter and so as to compensate this huge loss of heat, animals eat more during winter.
Numerical problems solution
- An aluminium can of mass 500 g contains 117.5 g of water at a temperature of 20 \(^0\) C. A 200 g block of iron at 75\(^0\) C is dropped into the can.
Find the final temperature, assuming no heat loss to the surrouding.
Solution:
mass of aluminium can, \(m_1\) = 500 g = 0.5 kg
mass of water, \(m_2\) = 117.5 g = 0.1175 kg
temperature of aluminium can and water, \(\theta_1\) = 20 \(^\circ\) C
mass of iron block, \(m_3\) = 200 g = 0.2 kg
temperature of iron block, \(\theta_2\) = 75\(^\circ\) C
specific heat capacity of aluminium, \(s_1\) = 882 J kg\(^{-1}\) \(^\circ\) C\(^{-1}\)
specific heat capacity of water, \(s_2\) = 4200 J kg\(^{-1}\) \(^\circ\) C\(^{-1}\)
specific heat capacity of iron, \(s_3\) = 470 J kg\(^{-1}\) \(^\circ\) C\(^{-1}\)
Let the final temperature be \(\theta\).
When the iron block is dropped into the aluminium can, heat is lost by the iron block and is gained by the can and the water.
Now,
Heat lost by the iron block, \[\begin{align*} &=m_3s_3 d\theta\\ &= 0.2 \times 470\times (75-\theta)\\ \end{align*}\] Heat gained by the aluminium can and the water is, \[\begin{align*} &=m_1 s_1 d\theta + m_2 s_2 d\theta \\ &=0.5 \times 882 \times (\theta-20) + 0.1175 \times 4200 \times (\theta-20) \end{align*}\] From the principle of calorimetry, \[\begin{align*} \text{heat lost}&=\text{heat gained}\\ 0.2 \times 470\times (75-\theta)&=0.5 \times 882 \times (\theta-20) + 0.1175 \times 4200 \times (\theta-20)\\ 7050-94\theta&=(441+493.5) \times (\theta-20)\\ 7050-94\theta&=934.5 \theta -18690\\ 7050+18690&=934.5 \theta+94\theta\\ 25740&=1028.5\theta\\ \theta&=25.02^\circ \hspace{0.1cm} \text{C} \end{align*}\] - A ball of copper(Specific heat capacity=400 J kg\(^{-1}\) k\(^{-1}\)) weighing 400 g transferred from a furnace to 1 kg of water at 20\(^\circ\) C.
The temperature of water rises to 50\(^\circ\) C. What is the original temperature of the ball?
Solution:
mass of copper ball, \(m_1\) = 400 g = 0.4 kg
mass of water, \(m_2\) = 1 kg
temperature of water, \(\theta_2\) = 20\(^\circ\) C
final temperature of water i.e., final temperature of the mixture, \(\theta\) = 50\(^\circ\) C
specific heat capacity of the copper, \(s_1\) = 400 J kg\(^{-1}\) k\(^{-1}\)
specific heat capacity of water, \(s_2\) = 4200 J kg\(^{-1}\) k\(^{-1}\)
original temperature of the ball, \(\theta_1\) = ?
When the copper ball is dropped into the water, the heat is gained by the water and is lost by the copper ball.
Now, heat lost by the copper ball is, \[\begin{align*} &=m_1s_1 d\theta\\ &=0.4 \times 400 \times (\theta_1 - 50)\\ &=160 \theta_1 - 8000 \end{align*}\] Heat gained by water is, \[\begin{align*} &=m_2 s_2 d\theta\\ &=1 \times 4200 \times (50-20)\\ &=126000 \hspace{0.1cm}\text{J} \end{align*}\] From the principle of calorimetry, \[\begin{align*} \text{heat lost}&=\text{heat gained}\\ 160 \theta_1 - 8000&=126000\\ 160 \theta_1&=134000\\ \theta_1&=837.5 ^\circ \hspace{0.1cm} \text{C} \end{align*}\] - A substance takes 3 min in cooling from 50 \(^0\) C to 45\(^0\) C and takes 5 min in cooling from 45\(^0\) C to 40\(^0\) C.
What is the temperature of the surroundings? How much time will it take to cool this substance from 40\(^0\) C to 35\(^0\) C?
Solution:
average temperature from 50 \(^\circ\) C to 45 \(^\circ\) C, \(\theta_1\) = \(\frac{50+45}{2}\) = 47.5 \(^\circ\) C
average temperature from 45 \(^\circ\) C to 40\(^\circ\) C, \(\theta_2\) = \(\frac{45+40}{2}\) = 42.5 \(^\circ\) C
time taken to cool from 50 \(^\circ\) C to 45 \(^\circ\) C, t\(_1\) = 3 min = 180 sec
time taken to cool from 45 \(^\circ\) C to 40 \(^\circ\) C, t\(_2\) = 5 min = 300 sec
temperature of the surrounding, \(\theta_0\) = ?
average temperature from 40 \(^\circ\) C to 35 \(^\circ\) C, \(\theta_3\) = 37.5 \(^\circ\) C
time taken to cool from 40 \(^\circ\) C to 35 \(^\circ\) C, t\(_3\) = ?
From Newton's law of cooling,
\[\begin{align*} \frac{dQ_1}{dt}&=-k(47.5-\theta_0)\\ ms\frac{d\theta_1}{dt}&=-k(47.5-\theta_0) ... (i)\\ \end{align*}\] And, \[\begin{align*} \frac{dQ_2}{dt}&=-k(42.5-\theta_0)\\ ms\frac{d\theta_2}{dt}&=-k(42.5-\theta_0) ... (ii)\\ \end{align*}\] Dividing eqn (i) by (ii), we get,
\[\begin{align*} \frac{\left(\frac{d\theta_1}{dt}\right)}{\left(\frac{d\theta_2}{dt}\right)}&=\frac{47.5-\theta_0}{42.5-\theta_0}\\ \frac{\left(\frac{50-45}{180}\right)}{\left(\frac{45-40}{300}\right)}&=\frac{47.5-\theta_0}{42.5-\theta_0}\\ \frac{5}{180}\times \frac{300}{5}&=\frac{47.5-\theta_0}{42.5-\theta_0}\\ \frac{5}{3}&=\frac{47.5-\theta_0}{42.5-\theta_0}\\ (42.5-\theta_0) \times 5&=(47.5-\theta_0)\times 3\\ 212.5-5\theta_0&=142.5-3\theta_0\\ 70&=2\theta_0\\ \theta_0&=35 \hspace{0.1cm} ^\circ \text{C} \end{align*}\] Also,
\[\begin{align*} \frac{dQ_3}{dt}&=-k(\theta_3-\theta_0)\\ ms\frac{d\theta_3}{dt}&=-k(37.5-35)... (iii)\\ \end{align*}\] Now, dividing equation (iii) by equation (ii), we get, \[\begin{align*} \frac{\frac{d\theta_3}{dt}}{\frac{d\theta_2}{dt}}&=\frac{2.5}{7.5}\\ \frac{\frac{40-35}{t_3}}{\frac{45-40}{300}}&=\frac{2.5}{7.5}\\ \frac{5}{t_3}\times \frac{300}{5}&=\frac{2.5}{7.5}\\ \frac{300}{t_3}&=\frac{2.5}{7.5}\\ 2.5t_3&=7.5 \times 300\\ t_3&=900 \hspace{0.1cm}\text{sec} \end{align*}\] So, it takes 15 mins to cool from 40\(^\circ\) C to 35\(^\circ\) C. - Determine the amount of heat required to convert 1 kg of ice at -10\(^0\) C to steam at 100\(^0\) C. Given:
Specific heat of ice=2100 J/Kg/K
Specific latent heat of ice= 336000 J/Kg
Specific latent heat of steam= 2260000 J/Kg}
Solution:
mass of ice, \(m_i\) = 1 kg
initial temperature = -10\(^0\) C
final temperature = 100\(^0\) C
Specific heat of ice,\(s_i\) =2100 J/Kg/K
Specific latent heat of ice, \(L_i\)= 336000 J/Kg
Specific latent heat of steam, \(L_s\)= 2260000 J/Kg
Heat required to convert ice from -10\(^0\) C to 0\(^0\) C ice is, \[\begin{align*} &=m_i \times s_i \times [0-(-10)]\\ &=1 \times 2100 \times 10\\ &= 21000 \hspace{0.1cm} \text{J} \end{align*}\] Heat required to convert ice from 0\(^0\) C to water at 0\(^0\) C is, \[\begin{align*} &=m_i \times L_i\\ &=1 \times 336000\\ &=336000 \hspace{0.1cm} \text{J} \end{align*}\] Heat required to convert water at 0\(^0\) C to water at 100\(^0\) C is, \[\begin{align*} &=m_i s_w (100-0)\\ &=1 \times 4200 \times 100\\ &=420000 \hspace{0.1cm} \text{J} \end{align*}\] Heat required to convert water at 100\(^0\) C to steam at 100\(^0\) C is, \[\begin{align*} &=m_iL_s\\ &= 1 \times 2260000\\ &=2260000 \hspace{0.1cm} \text{J} \end{align*}\] Thus, the total amount of heat required is, \[\begin{align*} &= 21000+336000+420000+2260000\\ &=3037000 \hspace{0.1cm} \text{J} \end{align*}\] - Find the result of mixing 20 g of water at 80\(^0\) C to 40 g of ice at -10\(^0\) C. (Specific heat of ice= 0.5 cal/g).
Solution:
( In this type of questions, you have to check whether all ice melts or not at first before proceeding. ) mass of water, \(m_w\) = 20 g
mass of ice, \(m_i\) = 40 g
temperature of water, \(\theta_w\) = 80\(^0\) C
temperature of ice, \(\theta_i\) = -10\(^0\) C
specific heat of ice, \(s_i\) = 0.5 cal/g
specific heat of water , \(s_w\) = 1 cal/g
Check!
Amount of heat required to convert 40 g of ice at -10\(^0\) C to ice at 0\(^0\) C is,
\[\begin{align*} &=m_i s_i [0-(-10)]\\ &=40\times 0.5 \times 10\\ &=200 \hspace{0.1cm}\text{cal} \end{align*}\] Amount of heat required to convert ice at 0\(^0\) C ice water at 0\(^0\) C is, \[\begin{align*} &=m_i L_i\\ &=40 \times 80\\ &=3200 \hspace{0.1cm}\text{cal} \end{align*}\] Amount of heat given out when 20 g water cools from 80 \(^\circ\) C to 0\(^\circ\) C is, \[\begin{align*} &=m_w s_w (80-0)\\ &=20 \times 1 \times 80\\ &=1600 \hspace{0.1cm}\text{cal} \end{align*}\] As the heat given out by water when it cools from 80 \(^\circ\) C to 0\(^\circ\) C i.e., 1600 \(\hspace{0.1cm}\text{cal}\) is less than that required to convert ice at -10 \(^\circ \) C to water at 0\(^\circ\) C i.e., 3400 cals, so all ice does not melt.
Proceed!
Heat available to melting of ice is, \[1600-200=1400 \hspace{0.1cm} \text{cals}\] Let, m mass of ice melts.
Then,
\[\begin{align*} mL_i&=1400\\ m\times 80&=1400\\ m&=17.5 \hspace{0.1cm}\text{g} \end{align*}\] Thus,
Resulting mass of ice = (40-17.5) = 22.5 g,
Resulting mass of water = (20+17.5) = 37.5 g
Resulting temperature of the mixture = 0\(^\circ\) C. - From what height must a hailstone at 0 degree celsius fall in order that it may melt on reaching the ground, assuming that
a) there is no loss of energy on the way b) 50 % of the energy is lost during fall.
Solution:
For an ice to melt, it requires heat energy equal to latent heat of fusion of ice.
So some of potential energy is retained by ice to melt.
In (a) there is no energy loss. So, all the potential energy is consumed in melting of ice. i.e., \[\begin{align*} mgh&=mL\\ gh&=L\\ 10\times h&=336000\\ h&=33600 \hspace{0.1cm}\text{m} \end{align*} \] So, for all ice to melt on reaching the ground assuming no energy loss, it must be thrown from the height of 33600 m.
In (b) 50 % of energy is lost during the fall. So, \[\begin{align*} 50\% \hspace{0.1cm}\text{of}\hspace{0.1cm} mgh&=mL\\ gh&=2L\\ 10 \times h&=2 \times 336000\\ h&=\frac{2\times 336000}{10}\\ h&=67200 \hspace{0.1cm}\text{m} \end{align*}\] So, for all ice to melt on reaching the ground when 50 % of the energy is lost, the ice must be thrown from the height of 67200 m. - 0.020 kg of ice and 0.10 kg of water at 0\(^0\) C are in a container. Steam at 100\(^0\) C is passed in until all the ice is just melted.
How much water is now in the container?
Specific heat of ice=2100 J/Kg/K
Specific latent heat of ice= 3.36\( \times 10^5\) J/Kg
Specific latent heat of steam= 2.2\( \times 10^6\) J/Kg
Specific heat capacity of water= 4.2 \(\times 10^3\) J/kg}
Solution:
Specific heat of ice, \(s_i\) = 2100 J/Kg/K
Specific latent heat of ice, \(L_i\) = 3.36\( \times 10^5\) J/Kg
Specific latent heat of steam, \(L_s\) = 2.2\( \times 10^6\) J/Kg
Specific heat capacity of water = \(s_w\) = 4.2 \(\times 10^3\) J/kg
Let, m mass of steam is condensed (i.e., changed into water).
Heat given out by steam in condensing and cooling to 0\(^\circ\) C (Remember! that all ice is just melted in our case and thus final temperature is 0\(^\circ\) C) is, \[\begin{align*} &=m L_s + m s_w (100-0)\\ &=m(2.2 \times 10^6+4200 \times 100)\\ &=m\times 2.62 \times 10^6 \end{align*}\] Heat absorbed by 0.02 kg of ice in melting is, \[\begin{align*} &=m_i L_i\\ &=0.020 \times 3.36\times 10^5\\ &=6720 \hspace{0.1cm} \text{J} \end{align*}\] From the principle of calorimetry, \[\begin{align*} \text{heat absorbed}&=\text{heat lost}\\ 6720&=m\times 2.62 \times 10^6\\ m&=0.0025 \hspace{0.1cm} \text{kg} \end{align*}\] So, total mass in the container is (0.10+0.020+0.0025) kg = 0.1225 kg.
Are you confused of why have i added 0.20 kg? It's because in addition to condensation of steam , the ice of 0.20 kg mass has also melted to water.
- 500 g of ice at -16\(^0\) C are dropped into a calorimeter containing 1000 g of water at 20\(^0\) C. The calorimeter is of copper and has
a mass of 278 g. Compute the final temperature of the system.
Specific heat capacity of copper = 400 J Kg\(^{-1}\) K\(^{-1}\)
Specific heat capacity of ice = 2100 J Kg\(^{-1}\) K\(^{-1}\)
Specific latent heat of ice = 336 \(\times 10^3\) J Kg\(^{-1}\) K\(^{-1}\)}Follow the same procedure as in q.no. 5
Solution:
mass of ice, \(m_i\) = 500 g = 0.5 kg
mass of water, \(m_w\) = 1000 g = 1 kg
mass of copper calorimeter, \(m_c\) = 278 g = 0.278 kg
temperature of ice, \(\theta_i\) = -16\(^0\) C
temperature of water, \(\theta_w\) = 20\(^0\) C
Specific heat capacity of ice, \(s_i\) = 2100 J Kg\(^{-1}\) K\(^{-1}\)
Specific latent heat of ice, \(L_i\) = 336 \(\times 10^3\) J Kg\(^{-1}\) K\(^{-1}\)
Specific heat capacity of copper, \(s_c\) = 400 J Kg\(^{-1}\) K\(^{-1}\)
Heat required to convert 0.5 kg of ice at -16\(^0\) C to ice at 0\(^\circ\) C is, \[\begin{align*} &=m_i s_i [0-(-16)]\\ &= 0.5 \times 2100 \times 16\\ &=16800 \hspace{0.1cm} \text{J} \end{align*}\] Heat required to convert 0.5 kg of ice at 0\(^\circ\) C to water at 0\(^\circ\) C is, \[\begin{align*} &=m_i L_i\\ &=0.5 \times 336 \times 10^3\\ &= 168000\hspace{0.1cm} \text{J} \end{align*}\] Heat given out by 1 kg of water and calorimeter as it cools from 20\(^\circ\) C to 0\(^\circ\) C is, \[\begin{align*} &=m_w s_w (20-0) + m_c s_c (20-0)\\ &=1 \times 4200 \times 20+0.278 \times 400 \times 20\\ &=84000+2224\\ &=86224\hspace{0.1cm} \text{J} \end{align*}\] So, amount of heat available for melting of ice is, \[86224-16800=69424\hspace{0.1cm} \text{J}\] Here, the heat given off by water is not sufficient enough to melt all the ice.
Since, the resulting mixture is a mixture of ice and water, the final temperature of the system is 0\(^\circ\) C.
Comments
Post a Comment