Question

A gold bar 20.0kg at 35.0°c is placed in a large insulated 0.8kg glass container at 15°c and 2.0kg of water at 25°c.. calculate the final equilibrium temperature?​

Answer 1

Answer:

The final equilibrium temperature is approximately 26.69 °C

Explanation:

The heat transferred, ΔQ, from a hot body to a cold one is given by the following formula;

ΔQ = m·c·ΔT

Where;

m = The mass of the body

c = The specific heat capacity of the body

ΔT = The temperature change of the body

The given mass of the gold bar, m₁ = 20.0 kg

The initial temperature of the gold bar, T₁ = 35.0 °C

The specific heat capacity of gold, c₁ = 0.13 kJ/(kg·K)

The mass of the glass container, m₂ = 0.8 kg

The initial temperature of the glass container, T₂ = 15°C

The specific heat capacity of glass, c₂ = 0.792 kJ/(kg·K)

The mass of the added water, m₃ = 2.0 kg

The initial temperature of the added water, T₃ = 25°C

The specific heat capacity of water, c₃ = 4.2 kJ/(kg·K)

The heat lost by the gold = The heat gained by the glass and the water

Let 'T' represent the temperature at the final equilibrium, we have;

m₁·c₁·ΔT₁ = m₂·c₂·ΔT₂ + m₃·c₃·ΔT₃

Where;

ΔT₁ = T₁ - T

ΔT₂ = T - T₂

ΔT₃ = T - T₃

∴ 20.0 × 0.13 × (35 - T) = 0.8 × 0.792 × (T - 15) + 2.0 × 4.2 × (T - 25)

Expanding and collecting like terms (using a graphing calculator) gives;

91 - 2.6·T = 9.0336·T - 219.504

9.0336·T + 2.6·T = 219.504 + 91 = 310.504

11.6336·T = 310.504

T = 310.504/11.6336 ≈ 26.69

The final equilibrium temperature, T ≈ 26.69 °C.