# Calorimetry

## Problems from IIT JEE

Problem (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.)

1. 8 min 20 second
2. 6 min 2 second
3. 7 min
4. 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}=\frac{(2)(4.2\times{10}^{3})(77-27)}{840}=500\;\mathrm{second}=8\;\mathrm{min}\;20\;\mathrm{second}.\nonumber \end{align}

Problem (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,

1. from $14.5\;\mathrm{{}^{o}C}$ to $15.5\;\mathrm{{}^{o}C}$ at 760 mm of Hg
2. from $98.5\;\mathrm{{}^{o}C}$ to $99.5\;\mathrm{{}^{o}C}$ at 760 mm of Hg
3. from $13.5\;\mathrm{{}^{o}C}$ to $14.5\;\mathrm{{}^{o}C}$ at 76 mm of Hg
4. 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 (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)}=\frac{(100)(1)(90-24)}{540+(1)(100-90)}=12\;\mathrm{g}.\nonumber \end{align}