Question

(II)A 2.3-kg lead ball is placed in a 2.5-L insulated pail of water initially at 20.0°C. If the final temperature of the water–lead combination is 32.0°C, what was the initial temperature of the lead ball?

Short answer

The initial temperature of the lead ball is ${452}^{\circ }\mathrm{C}$.

Step-by-step solution

Given data

The mass of the lead ball is $m=2.3\mathrm{k}\mathrm{g}$.

The amount of water is $m^{\prime }=2.5\mathrm{L}$.

The temperature of the water is $T={20}^{\circ }\mathrm{C}$.

The temperature of the water-lead combination is $T^{\prime }={32}^{\circ }\mathrm{C}$.

Understanding the heat lost by the lead

In this problem, use the relation of heat transfer to calculate the heat lost by the lead. Also, the heat loss is equal to the heat gained by the insulated pail of water.

Calculation of the initial temperature of the ball

The relation to calculate the temperature is given by:

$\begin{array}{r}c{Q}_{\mathrm{L}}=Q_{\mathrm{w}}\\ mc\left(T_{\mathrm{i}}-T^{\prime }\right)=m^{\prime }{c}^{\prime }\left(T^{\prime }-T\right)\end{array}$

Here,$c$and$c^{\prime }$are the specific heat of lead and water, and$T_{\mathrm{i}}$is the initial temperature of the lead ball.

On plugging the values in the above relation, you get:

$\begin{array}{r}c\left(2.3\mathrm{k}\mathrm{g}\right)\left(130\mathrm{J}/\mathrm{k}\mathrm{g}{\cdot }^{\circ }\mathrm{C}\right)\left(T_{\mathrm{i}}-{32}^{\circ }\mathrm{C}\right)=\left(2.5\mathrm{L}\times \frac{1\mathrm{k}\mathrm{g}}{1\mathrm{L}}\right)\left(4186\mathrm{J}/\mathrm{k}\mathrm{g}{\cdot }^{\circ }\mathrm{C}\right)\left({32}^{\circ }\mathrm{C}-{20}^{\circ }\mathrm{C}\right)\\ T_{\mathrm{i}}={452}^{\circ }\mathrm{C}\end{array}$

Thus, $T_{\mathrm{i}}={452}^{\circ }\mathrm{C}$ is the required initial temperature.