Calorimetry
By Calorimetry deals with the measurement of the heat energy absorbed or released by a system during a process. When two bodies at different temperatures come in contact, heat is transferred from the body at a higher temperature to the body at a lower temperature. According to the principle of calorimetry (energy conservation), heat loss is equal to the heat gained. The heat gained by a body is related to the rise in its temperature $\Delta T$ by \begin{align} Q=ms\Delta T \end{align} where $m$ is the mass of the body and $s$ is the specific heat of the material of the body.
When there is a change of state (phase transitions like fusion and vaporization), the body gain or lose heat without a change in temperature. The amount of heat transfer is given by \begin{align} Q=mL \end{align} where $L$ is latent heat of the material of the body. For water, the latent heat of fusion is 80 cal/g and the latent heat of vaporization is 540 cal/g.
The work done on a body can be used to increase its temperature. The work (in J) is related to the heat (in cal) by $Q=WJ$, where $J=4.18$ cal/g is called the mechanical equivalent of heat.
A calorimeter is used to measure the amount of heat transferred when two bodies are brought in contact. The water equivalent of a calorimeter is the mass of water that would need the same quantity of heat transfer as the calorimeter to create the same temperature change. Its unit is kg.
Mixing Two Substances Without Change of State: Consider mixing of two substances of masses $m_1$ and $m_2$, specific heats $s_1$ and $s_2$ and temperatures $T_1$ and $ T_2 \,( < T_1 ) $. Let temperature of the mixture be $T$. By principle of calorimetry, \begin{align} m_1 s_1 (T_1-T)=m_2 s_2 (T-T_2). \end{align} Solve to get temperature of the mixture at equilibrium as \begin{align} T=\frac{m_1 s_1 T_1+m_2 s_2 T_2}{m_1 s_1 +m_2 s_2} \end{align}
Mixing Two Substances With Change of State: Suppose water at temperature $T_w$ deg C is mixed with ice at 0 deg C. In this case, first ice will melt and then its temperature will rise to attain thermal equilibrium. It is important to consider the latent heat during phase transition.
Solved Problems on Calorimetry
Problem from IIT JEE 2005
Water of volume 2 litre in a container is heated with a coil of 1 kW at $27\;\mathrm{{}^{o}C}$. The lid of the container is open and energy dissipates at rate of 160 J/s. In how much time temperature will rise from $27\;\mathrm{{}^{o}C}$ to $77\;\mathrm{{}^{o}C}$? (Specific heat of water is 4.2 kJ/kg.)
- 8 min 20 second
- 6 min 2 second
- 7 min
- 14 min
Solution: The heater coil gives energy at a rate of 1000 J/s, out of which 160 J/s is dissipated through the lid. Thus, energy for heating the water is available at a rate of, \begin{align} \label{xpa:eqn:1} P=1000-160=840\; \mathrm{J/s}. \end{align} The heat energy required to raise the temperature of mass $m$ of water from $T_1$ to $T_2$ is, \begin{align} \label{xpa:eqn:2} Q=mS(T_2-T_1), \end{align} where $S$ is specific heat of the water. If $t$ is the time required to raise temperature of mass $m$ from $T_1$ to $T_2$ then, by energy conservation, \begin{align} \label{xpa:eqn:3} Pt=Q. \end{align} The density of water is $1000\;\mathrm{kg/m^3}$. Thus, mass of 2 litre of water is $m=2\;\mathrm{kg}$. Substitute $P$ and $Q$ from first and second equations into third equation to get, \begin{align} t&=\frac{mS(T_2-T_1)}{P}\nonumber\\ &=500\;\mathrm{sec}=8\;\mathrm{min}\;20\;\mathrm{sec}.\nonumber \end{align}
Problem from IIT JEE 2005
One calorie is defined as the amount of heat required to raise temperature of 1 g of water by $1\;\mathrm{{}^{o}C}$ in a certain interval of temperature and at certain pressure. The temperature interval and pressure is,
- from $14.5\;\mathrm{{}^{o}C}$ to $15.5\;\mathrm{{}^{o}C}$ at 760 mm of Hg
- from $98.5\;\mathrm{{}^{o}C}$ to $99.5\;\mathrm{{}^{o}C}$ at 760 mm of Hg
- from $13.5\;\mathrm{{}^{o}C}$ to $14.5\;\mathrm{{}^{o}C}$ at 76 mm of Hg
- from $3.5\;\mathrm{{}^{o}C}$ to $4.5\;\mathrm{{}^{o}C}$ at 76 mm of Hg
Solution: One calorie is defined as the heat required to raise temperature of 1 g of water from $14.5\;\mathrm{{}^{o}C}$ to $15.5\;\mathrm{{}^{o}C}$ at atmospheric pressure.
Problem from IIT JEE 1996
The temperature of 100 g of water is to be raised from $24\;\mathrm{{}^{o}C}$ to $90\;\mathrm{{}^{o}C}$ by adding steam to it. Calculate the mass of the steam required for this purpose.
Solution: Let $m_s$ be mass of the steam required to raise the temperature of mass $m_w=100\;\mathrm{g}$ of water from temperature $T_1=24\;\mathrm{{}^{o}C}$ to temperature $T_2=90\;\mathrm{{}^{o}C}$. In this process, mass $m_s$ of steam at temperature $T_s=100\;\mathrm{{}^{o}C}$ is transformed to water at temperature $T_s$ by releasing heat, \begin{align} \label{dsa:eqn:1} Q_\text{r,1}=mL. \end{align} Then, mass $m_s$ of water at temperature $T_s$ is cooled to temperature $T_2$ by releasing heat, \begin{align} \label{dsa:eqn:2} Q_\text{r,2}=mS_w(T_s-T_2). \end{align} The mass $m_w$ of water is heated from temperature $T_1$ to temperature $T_2$ by absorbing heat, \begin{align} \label{dsa:eqn:3} Q_\text{a}=m_w S_w (T_2-T_1). \end{align} By energy conservation, $Q_\text{r,1}+Q_\text{r,2}=Q_\text{a}$. Using above equations, we get, \begin{align} m_s=\frac{m_w S_w(T_2-T_1)}{L+S_w(T_s-T_2)}=12\;\mathrm{g}.\nonumber \end{align}
Problem from IIT JEE 2003
2 kg of ice at -20 deg C is mixed with 5 kg of water at 20 deg C in an insulating vessel having a negligible heat capacity. Calculate the final mass of water remaining in the container. It is given that the specific heats of water and ice are 1 kcal/kg-C and 0.5 kcal/kg-C while the latent heat of fusion of ice is 80 kcal/kg.
- 7 kg
- 6 kg
- 4 kg
- 2 kg
Solution: The heat energy absorbed, when 2 kg of ice at -20 deg C is heated to ice at 0 deg C, \begin{align} Q_\text{ice}&=m_iS_i\Delta T\nonumber\\ &=(2)(0.5)(20)=20\, \mathrm{kcal}. \nonumber \end{align} The heat energy released, when 5 kg of water at 20 deg C is cooled to water at 0 deg C, \begin{align} Q_\text{water}&=m_wS_w\Delta T \nonumber\\ &=(5)(1) (20)=100\,\mathrm{kcal}. \nonumber \end{align} Thus, the energy available to melt the ice is \begin{align} Q&=Q_\text{water}-Q_\text{ice}\nonumber\\ &=100-20=80\,\mathrm{kcal}, \nonumber \end{align} and the mass of ice melted by this energy is \begin{align} m=Q/L_\text{fusion}=1\, \mathrm{kg}. \nonumber \end{align} Hence, the final mass of water in the container is $m_w+m=6\,\mathrm{kg}$. We encourage you to prove that the final state of matter cannot be (i) only water or (ii) only ice. Hint: These states violate energy conservation.
Problem from IIT JEE 1986
Steam at 100 deg C is passed into 1.1 kg of water contained in a calorimeter of water equivalent 0.02 kg at 15 deg C till the temperature of the calorimeter and its contents rises to 80 deg C. The mass of the steam condensed (in kg) is
- 0.130
- 0.065
- 0.260
- 0.135
Solution: The heat required to raise the temperature of the water of mass $m_w=1.1\,\mathrm{kg}$ and calorimeter of water equivalent $m_c=0.02\,\mathrm{kg}$ from initial temperature $T_i=15$ deg C to final temperature 80 deg C is given by \begin{align} Q_1&=(m_w+m_c)S(T_f-T_i)\nonumber\\ &=(1.1+0.02)(1000)(80-15)\nonumber\\ &=72800\,\mathrm{cal}. \nonumber \end{align} The heat released in the condensation of mass $m_s$ of steam at 100 deg C is \begin{align} Q_2=m_sL=m_s (540000)\,\mathrm{cal}. \nonumber \end{align} The energy conservation, $Q_1=Q_2$, gives $m_s=72800/540000=0.135\,\mathrm{kg}=135\,\mathrm{g}$. We encourage you to find $m_s$, if the final temperature of the condensed steam is reduced to 80 deg C. Hint: $m_s=130\mathrm{g}$.
A Practical Problem (Cooling of Rooftop by Water Vaporization)
Problem: In summers, the rooftop becomes very hot due to incident sunlight. This increases the temperature inside the house. One of the solutions is to sprinkle water on the rooftop. The vaporizing water takes away heat from the roof and cools it down. Let us estimate the amount of water required to cool the rooftop. The concrete roof is of size $4m\times4m\times10cm$. The density of concrete is $6000$\cubicm and its specific heat is $880$J/(kg-deg C). Assume that the vaporization of water takes away heat from the roof uniformly. How much water is required to reduce the roof temperature by $4$deg C? The latent heat of vaporization of water is $2.25\times{10}^{6}$J/kg.
- $10$kg
- $15$kg
- $20$kg
- $30$kg
Solution: The volume of the roof is $V=4\times4\times0.1=1.6$ m3 and its mass is \begin{align} M &=\rho V=6000\times1.6 \\ &=9.6\times{10}^{3}\;\mathrm{kg}.\nonumber \end{align} The amount of heat to be taken away to reduce roof temperature by $\Delta T=4$ deg C is \begin{align} Q&=MS\Delta T \\ &=(9.6\times10^3)(880)(4)\\ &=2.38\times10^7\;\mathrm{J}.\nonumber \end{align} Let the required mass of the water be $m$. The heat $Q$ shall be equal to the heat required for vaporization of water i.e., $Q=mL$, which gives \begin{align} m&=\frac{Q}{L} \\ &=\frac{2.38\times10^7}{2.25\times10^6} \\ &=15\;\mathrm{kg}.\nonumber \end{align}
A Practical Problem (Thermodynamics of Making Tea)
Problem: A tea-making utensil containing 500 g water is placed on a burner. The water takes 5 minutes to starts boiling. If the initial temperature of the water is 40 deg C and the water equivalent of the utensil is 20 g then the burner supply heat to the utensil at a rate of
- 100 cal/s
- 104 cal/s
- 105 cal/s
- 110 cal/s
Solution: The specific heat of water is 1 cal/g, and its boiling point is $T_b=100$ deg C. The effective mass of the water is $m=500+20=520$ g. The total heat supplied by the burner in 5 minutes is \begin{align} Q&=mS(T_b-T_0)=(520)(1)(100-40)\nonumber\\ &=520\times60\;\mathrm{cal}.\nonumber \end{align} Thus, heat supplied by the burner per second is \begin{align} \frac{\mathrm{d}Q}{\mathrm{d}t}&=\frac{Q}{t}\\ &=\frac{520\times 60}{5\times 60}\\ &=104 \;\mathrm{cal/s}.\nonumber \end{align}
Questions on Calorimetry
Question 1: If the specific heat of a substance is infinite, it means
Question 2: The amount of heat required to convert 1 gm of ice at 0 deg C into steam at 100 deg C is
