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QPage 47: Q1: HC Verma: Concepts of Physics - Volume 2 Calorimetry Page 47: Q1 Solution: An aluminium vessel of mass 0.5 kg contains 0.2 kg of water at…

⚠️ Common Exam Traps & Conceptual Pitfalls

Students frequently fail to explicitly define the temperature difference as (Final - Initial) for heat gain and (Initial - Final) for heat loss, leading to sign errors in the energy balance equation. Always ensure that the algebraic signs reflect the physical direction of heat flow, such that heat lost by the hot body equals heat gained by the cold bodies.

An aluminium vessel of mass 0.5 kg contains 0.2 kg of water at 20�C. A block of iron of mass 0.2 kg at 100�C is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron and water are 910 J kg−1 K−1910 \text{ J kg}^{-1} \text{ K}^{-1}, 470 J kg−1 K−1470 \text{ J kg}^{-1} \text{ K}^{-1} and 4200 J kg−1 K−14200 \text{ J kg}^{-1} \text{ K}^{-1} respectively.

Step-by-Step Verified Solution

**Answer:** The equilibrium temperature of the mixture is 25�C. **Step-by-Step Explanation:** Given: Mass of aluminium ma=0.5m_a = 0.5 kg; Mass of water mw=0.2m_w = 0.2 kg; Mass of iron mi=0.2m_i = 0.2 kg; Temperature of aluminium and water = 20�C = 293 K; Temperature of iron = 100�C = 373 K; Specific heat of aluminium = 910 J/kg-K; Specific heat of iron = 470 J/kg-K; Specific heat of water = 4200 J/kg-K. Assumption: heat interaction outside of the boundary is zero. Heat loss by iron block = heat gain by aluminium vessel and water
0.2×470×(373−T)=(T−293)(0.5×910+0.2×4200)0.2 \times 470 \times (373 - T) = (T - 293)(0.5 \times 910 + 0.2 \times 4200)
94(373−T)=(T−293)(455+840)94(373 - T) = (T - 293)(455 + 840)
373−T=(T−293)×13.8373 - T = (T - 293) \times 13.8
T=298 K=25�CT = 298 \text{ K} = 25�\text{C}
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