Question

A 0.225 kg sample of tin initially at 97.5°c is dropped into 0.115 kg of water. the initial temperature of the water is 10.0°c. if the specific heat capacity of tin is 230 j/kg • °c, what is the final equilibrium temperature of the tin-water mixture

Answer 1

Answer:

The final equilibrium temperature of the tin-water mixture is approximately 18.468 °C

Explanation:

The parameters of heat energy transfer from the tin to the water are given as follows;

The mass of the sample of tin, m₁ = 0.225 kg

The initial temperature of the tin, T₁ = 97.5 °C

The mass of the water into which the tin is dropped, m₂ = 0.115 kg

The initial temperature of the water, T₂ = 10.0 °C

The specific heat capacity of tin, c₁ = 230 J/(kg·°C)

The specific heat capacity of water, c₂ = 4,200 J/(kg·°C)

Let 'T' represent the final equilibrium temperature of the tin-water mixture, we have;

The heat lost by the tin, ΔQ$_{tin}$ = The heat gained by the water ΔQ$_{water}$

∴  ΔQ$_{tin}$ = ΔQ$_{water}$

Where;

ΔQ$_{tin}$ = m₁·c₁·(T₁ - T)

ΔQ$_{water}$ = m₂·c₂·(T - T₂)

By substitution, we have;

ΔQ$_{tin}$ = 0.225 kg × 230 J/(kg·°C) × (97.5°C - T)

ΔQ$_{water}$ = 0.115 kg × 4,200 J/(kg·°C) × (T - 10.0°C)

From ΔQ$_{tin}$ = ΔQ$_{water}$, we have;

0.225 kg × 230 J/(kg·°C) × (97.5°C - T) = 0.115 kg × 4,200 J/(kg·°C) × (T - 10.0°C)

∴ 5,045.625 J - 51.75 J/°C × T = 483 J/°C × T - 4,830 J

5,045.625 J + 4,830 J = 534.75 J/°C × T

∴ 534.75 J/°C × T = 9,875.625 J

T =  9,875.625 J/(534.75 J/°C) = 18.4677419 °C ≈ 18.468 °C

The final equilibrium temperature of the tin-water mixture, T ≈ 18.468 °C.