Chapter 21, Problem 66A

A cook pours L of ice water at $0°C$ into a pan of hot water at $80°C$ and finds that the mixture reaches a temperature of $60°C$. How much hot water was in the pan?

To determine

Amount of hot water in the pan.

Answer to Problem 66A

Amount of hot water in the pan is, $V_2=3\text{ L}$

Explanation of Solution

Given :

Volume of ice water at $0°\text{C}$ is, $V_1=1\text{ L}$

Temperature of hot water, $T_2=80°\text{ C}$

Temperature of mixture, $T=60°\text{C}$

Formula used :

  $Q=mc\Delta t$

Where, Q is amount of heat generated, m is mass, c is specific heat and $\Delta t$ is change in temperature.

Calculation :

Density of water is, $d=1000{\text{ kg/m}}^3$

Volume of cold water, $V_1=1\text{ L}={10}^{-3}{\text{ m}}^3$

Therefore, mass of cold water,

  $\begin{array}{c}{m}_1=d{V}_1\\ =1000\times {10}^{-3}\\ =1\text{ kg}\end{array}$

Let mass of hot water is $m_2\text{ kg}$

According to heat transfer heat lost by hot water is equal to heat gained by cold water.

  $m_{1}c\Delta t_1=m_{2}c\Delta t_2$

  $\begin{array}{c}1\times c\times \left(60-0\right)=m_2\times c\times \left(80-60\right)\\ 60=20m_2\\ 20m_2=60\\ m_2=\frac{60}{20}\\ =3\text{ kg}\end{array}$

Density of water is, $d=1000{\text{ kg/m}}^3$

Therefore, volume of hot water,

  $\begin{array}{c}{V}_2=\frac{3\text{ kg}}{1000{\text{ kg/m}}^3}\\ =3\times {10}^{-3}{\text{ m}}^3\\ =3\text{ L}\end{array}$

Conclusion :

Therefore, 3 L hot water was in the pan.