Chapter 12, Problem 44A

To determine

The amount of heat absorbed by the coffee cup.

Answer to Problem 44A

  $2.02\times {10}^4\text{J}$

Explanation of Solution

Given:

Mass $m=4\times {10}^2\text{g = 0}\text{.4 kg}$

Specific heat capacity of glass, $c=\text{840}\frac{\text{J}}{\text{kg}\cdot °\text{C}}$

Temperature, ${\text{T}}_i=20°\text{C}$ and ${\text{T}}_f=\text{80}°\text{C}$

Change in temperature, $\Delta \text{T}=6\text{0}°\text{C}$

Formula Used:

The heat absorbed or released is

  $Q=mc\Delta T$

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

Calculation:

Substituting values.

  $Q=\left(\text{0}\text{.4}\text{kg}\right)\left(\text{840 }\frac{\text{J}}{\text{kg}\cdot °\text{C}}\right)\left(6\text{0}°\text{C}\right)$

Simplifying

  $Q=2.02\times {10}^4\text{J}$

Conclusion:

The amount of heat absorbed by the coffee cupis $2.02\times {10}^4\text{J}$