Chapter 20, Problem 21SP

A 2.0 kg metal block (c = 0.137 cal/g • °C) is heated from 15 °C to 90 °C. By how much does its internal energy change?

To determine

The change in internal energy of $2.0\text{kg}$ metal block if it is heated from 15 °C to 90 °C and its specific heat is $0.137\text{cal/g}\cdot \text{°C}$.

Answer to Problem 21SP

Solution:

$86\text{ kJ}$

Explanation of Solution

Given data:

Mass of the metal block is $2.0\text{kg}$.

Temperature changes from $15°\text{C}$ to $90°\text{C}$.

Specific heat of the metal block is $0.137\text{ cal/g}\cdot °\text{C}$.

Formula used:

First law of thermodynamics:

$\Delta Q=\Delta U+\Delta W$

Here, $\Delta Q$ is the heat added or removed to the system, $\Delta U$ is the internal energy, and $\Delta W$ is the work done by or on the system.

Write the expression for heat gain or loss if a body or substance undergoes temperature changes;

$\Delta Q=mc\Delta T$

Here, $m$ is the mass of the substance, $c$ is the specific heat of the substance, and $\Delta T$ is the change in temperature.

Explanation:

Calculate the change in temperature:

$\begin{array}{c}\Delta T=90°\text{C}-15°\text{C}\\ =75°\text{C}\end{array}$

Recall the expression for heat required to raise the temperature:

$\Delta Q=mc\Delta T$

Substitute $2.0\text{kg}$ for $m$, $75°\text{C}$ for $\Delta T$, and $0.137\text{ cal/g}\cdot °\text{C}$ for $c$.

$\begin{array}{c}\Delta Q=\left(2.0\text{kg}\right)\left(0.137\text{ cal/g}\cdot °\text{C}\right)\left(75°\text{C}\right)\\ =\left(2.0\text{kg}\right)\left(\frac{1000\text{g}}{1\text{ kg}}\right)\left(0.137\text{ cal/g}\cdot °\text{C}\right)\left(\frac{4.18\text{ J}}{1\text{ cal}}\right)\left(75°\text{C}\right)\\ \approx 8.6\times {10}^3\text{J}\left(\frac{1\text{kJ}}{1000\text{ J}}\right)\\ =86\text{ kJ}\end{array}$

There isnot any movement in the metal block. Therefore, the work done by or on the system is zero.

Recall the expression of first law of thermodynamics:

$\Delta Q=\Delta U+\Delta W$

Substitute $86\text{ kJ}$ for $\Delta Q$, and $0\text{ kJ}$ for $\Delta W$

$\begin{array}{c}86\text{ kJ}=\Delta U+0\\ \Delta U=86\text{ kJ}\end{array}$

Conclusion:

The change in internal energy of the given metal block is $86\text{ kJ}$.