**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.)

- 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}=\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,

- 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 (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}