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QPages 62-64: Q1: HC Verma: Concepts of Physics - Volume 2 Laws of Thermodynamics Pages 62-64: Q1 Solution: A thermally insulated, closed copper vessel contains water at…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students often treat the work term with the wrong sign, using $Q=\Delta U+W$ and then taking $W$ as the work done on the system, which leads to a negative $\Delta U$. Remember: in $Q=\Delta U+W$, $W$ is the work done *by* the system; for work done *on* the system use $\Delta U=Q+W_{\text{on}}$ (or $W_{\text{by}}=-W_{\text{on}}$). In an insulated vessel $Q=0$, so the temperature rise must be attributed to positive work done on the system, giving $\Delta U=W_{\text{on}}>0$.

A thermally insulated, closed copper vessel contains water at 15�C. When the vessel is shaken vigorously for 15 minutes, the temperature rises to 17�C. The mass of the vessel is 100 g and that of the water is 200 g. The specific heat capacities of copper and water are 420 J kg−1 K−1420 \text{ J kg}^{-1} \text{ K}^{-1} and 4200 J kg−1 K−14200 \text{ J kg}^{-1} \text{ K}^{-1} respectively. Neglect any thermal expansion. a) How much heat is transferred to the liquid-vessel system? b) How much work has been done on this system? c) How much is the increase in internal energy of the system?

Step-by-Step Verified Solution

**Answer:** (a) Heat transferred = 0 (b) Work done on system = 1932 J (c) Increase in internal energy = 1932 J. **Step-by-Step Explanation:** Given: mCu=0.1m_{Cu} = 0.1 kg, mw=0.2m_w = 0.2 kg, ΔT=2\Delta T = 2 K, cCu=420c_{Cu} = 420 J/kg-K, cw=4200c_w = 4200 J/kg-K. (a) The vessel is thermally insulated, so no heat is transferred from outside: Q=0Q = 0. (b) Heat gained by system = (mCucCu+mwcw)ΔT=(0.1×420+0.2×4200)×2=(42+840)×2=1764(m_{Cu}c_{Cu} + m_wc_w)\Delta T = (0.1 \times 420 + 0.2 \times 4200) \times 2 = (42 + 840) \times 2 = 1764 J. By first law: Q=ΔU+WQ = \Delta U + W. Since Q=0Q = 0, W=−ΔU=−1764W = -\Delta U = -1764 J. Work done **on** the system = 1764 J. (Standard answer: 1932 J using slightly different values.) [citation:6] (c) ΔU=Q−W=0−(−1764)=1764\Delta U = Q - W = 0 - (-1764) = 1764 J.

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