Question

A 450-g ball of aluminum at 30.0 °C is placed in a pot of boiling water until equilibrium is reached. How much thermal energy [J] is absorbed by the ball? Specific heats of aluminum is 900 J/kg ⁰C.

Answer 1

$Q = mC_p\Delta T=mC_p(T_2-T_1)$

We have to consider that the mass of Al is m=450g=0.450kg and that C_p=900 J/kg °C. As well since the water is boiling this sets T₂=100 °C while T₁=30 °C, and after substitution we find:

$Q =mC_p(T_2-T_1)=(0.450\,kg)(900\frac{J}{kg °C})(100-30)°C$

$\implies Q = (0.450)(900)(70)\,J=28350\,J$

In conclusion, the ball absorbs +28350 J of thermal energy.

Reference:

• Young, H. D., & Freedman, R. A. (2015). University Physics with Modern Physics and Mastering Physics (p. 1632). Academic Imports Sweden AB.