Question

A 10 kg solid at 100°C with a specific heat, of 0.8 kJ/kg°C is immersed in 40 kg of 20°C liquid with a specific heat of 4.0 kJ/kg°C. Estimate the temperature after a long time if the container is insulated?
1. 30°C
2. 28°C
3. 26°C
4. 24°C

Answer 1

Correct Answer - Option 4 : 24°C

Concept:

The heat transfer by solid is equal to the heat gain by liquid...........(as container is insulated)

Q_s = Q_L

(m × C × ΔT)_s = (m × C × ΔT)_L

Calculation:

Given:

m_s = 10 kg, T_s = 100°C, C_s = 0.8 kJ/kg°C

m_L = 40 kg, T_L = 20°C, C_L = 4.0 kJ/kgK

Let T_f is the intermediater temperature between solid and liquid,

Therefore, 10 × 0.8 × (100 – T_f) = 40 × 4.0 × (T_f – 20)

\(\Rightarrow \frac{{\left( {100 - {T_f}} \right)}}{{\left( {{T_f} - 20} \right)}} = \frac{{40 × 4.0}}{8} = 20\)

100 – T_f = 20T_f – 400

21T_f = 500

⇒ \(Tf = \frac{{500}}{{21}}\) = 23.8 ≈ 24° C