Specific-Heat and Calorimetry Calculations: 22-Agric-A3 Heat Engineering, December 2017, Q2

Problem 2: Specific-Heat and Calorimetry Calculations (25 points)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
(a) Tea volume / density / specific heat	200 cm³ / 1000 kg/m³ / 4186 J/kg·C
(a) Tea initial temperature	95°C
(a) Cup mass / specific heat / initial temperature	0.150 kg / 840 J/kg·C / 25°C
(b) Aluminum mass / specific heat / temperature rise	1 kg / 900 J/kg·K / 30°C → 100°C
(c) Solid mass / temperature rise / heat transferred	2 kg / 300 K → 550 K / 50 kJ

Find. (a) The equilibrium mixture temperature T; (b) the heat Q needed to raise the aluminum block's temperature; (c) the specific heat c of the 2 kg solid.

Approach. Each part is a direct application of sensible-heat accounting, $Q = mc\,\Delta T$: part (a) balances heat lost by the tea against heat gained by the cup (a closed, adiabatic mixing system, so the two sides are set equal); parts (b) and (c) apply the same relation once forward (solve for Q) and once inverted (solve for c).

Part (a) — equilibrium temperature of the tea/cup mixture. The tea's mass follows from its volume and density, $m_{tea}=\rho V = 1000\times200\times10^{-6}=0.200\ \text{kg}$. With no heat lost to the surroundings, heat lost by the tea equals heat gained by the cup: $$m_{tea}c_{tea}(T_{tea}-T) = m_{cup}c_{cup}(T-T_{cup})$$ $$0.200(4186)(95-T) = 0.150(840)(T-25)$$ Solving for T, $$T = \frac{m_{tea}c_{tea}T_{tea}+m_{cup}c_{cup}T_{cup}}{m_{tea}c_{tea}+m_{cup}c_{cup}} = \frac{(837.2)(95)+(126)(25)}{837.2+126} = \boxed{85.8\,{}^{\circ}\text{C}}$$
Part (b) — heat to raise the aluminum block's temperature. Direct sensible-heat accounting, $$Q = mc\,\Delta T = 1\ \text{kg}\times900\ \frac{\text{J}}{\text{kg}\cdot\text{K}}\times (100-30)\ \text{K} = \boxed{63{,}000\ \text{J} = 63\ \text{kJ}}$$
Part (c) — specific heat of the solid. The reservoir temperature (1000 K) fixes the direction of heat flow but does not enter the specific-heat calculation itself — only the heat actually absorbed by the block and its own temperature rise do. Inverting $Q=mc\,\Delta T$, $$c = \frac{Q}{m\,\Delta T} = \frac{50{,}000\ \text{J}}{2\ \text{kg}\times(550-300)\ \text{K}} = \boxed{100\ \text{J/kg}\cdot\text{K}}$$

Quantity	Result
(a) Equilibrium mixture temperature, T	85.8°C
(b) Heat to raise the aluminum block, Q	63,000 J (63 kJ)
(c) Specific heat of the solid, c	100 J/kg·K

Concept & Theory

Topic: Sensible-heat accounting and calorimetric mixing.

Key relations: $Q = mc\,\Delta T$; conservation of energy for an adiabatic mixing process, $\sum Q_{lost} = \sum Q_{gained}$, i.e. $m_1c_1(T_1-T)=m_2c_2(T-T_2)$.

Why this works: With the system isolated from the surroundings, the First Law reduces to internal energy exchange between the two bodies only; because both stay well below any phase-change temperature, each body's internal-energy change is captured by its sensible-heat term $mc\,\Delta T$, and the two terms must sum to zero (heat lost by the hotter body equals heat gained by the cooler one). Rearranging the same relation for the unknown specific heat in part (c) is legitimate because $Q$, $m$, and $\Delta T$ are all independently given — the reservoir's own temperature is extraneous data for that sub-question (it only matters if a second-law/entropy-generation question is asked about the transfer).

Common pitfall: Mixing up which side of the balance is "lost" vs. "gained" (the tea must cool towards the cup's temperature, not the reverse), and treating the reservoir temperature in part (c) as if it belonged in the specific-heat formula.

Source: Çengel & Ghajar, Heat and Mass Transfer, Ch. 1 (sensible heat, specific heat, closed-system energy balance).

Question 2 of 4: Specific-Heat and Calorimetry Calculations

Problem 2: Specific-Heat and Calorimetry Calculations (25 points)