Thermal Physics: Practice Problem & Solution
The internal energy change in a system that has absorbed $2\text{ kcal}$ of heat and done $500\text{ J}$ of work is: (2009)
Solution Explained:
To solve this problem, we apply the core principles of Thermal Physics. Understanding the underlying formula is key to arriving at the correct answer below:
Using $\Delta Q = 2\text{ kcal} = 2000 \times 4.2\text{ J} = 8400\text{ J}$ and $\Delta W = 500\text{ J}$. From the first law, $\Delta U = \Delta Q - \Delta W = 8400 - 500 = 7900\text{ J}$.
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