Question

Specific heat capacity of water = 4200 J/kg/K
Specific heat capacity of aluminium = 900 J/kg/°C
Specific heat capacity of ice = 2100 J/ kg/K
Latent heat of fusion of ice = 334 000 J/ kg
Latent heat of vaporization of water =
2 250 000J/kg

We wish to determine the specific heat capacity of a new alloy of an unknown specific heat capacity.

A 0.15 kg sample of alloy is heated to 540°C. It is then quickly placed in 400 g of water at l0°C which is contained in a 200 g aluminium cup. The final temperature of the mixture is 305°C. Calculate the specific heat capacity of the alloy.

NB: Heat lost by alloy is gained by both the water and the cup containing the water.

Answer 1

Answer on Question #80209, Physics / Molecular Physics | Thermodynamics

Specific heat capacity of water = 4200 J/kg/K

Specific heat capacity of aluminium = 900 J/kg/°C

Specific heat capacity of ice = 2100 J/kg/K

Latent heat of fusion of ice = 334 000 J/kg

Latent heat of vaporization of water = 2 250 000 J/kg

We wish to determine the specific heat capacity of a new alloy of an unknown specific heat capacity.

A 0.15 kg sample of alloy is heated to 540°C. It is then quickly placed in 400 g of water at 10°C which is contained in a 200 g aluminium cup. The final temperature of the mixture is 305°C. Calculate the specific heat capacity of the alloy.

NB: Heat lost by alloy is gained by both the water and the cup containing the water.

Solution

The lost heat is equal to the gained heat.

$$Q_{\text{lost}} = Q_{\text{gained}}$$

The sample $f$ alloy is the one that loses heat:

$Q_{\text{lost}} = c_{\text{al}} m \Delta T = c_{\text{al}} m (T_2 - T_1)$, where $T_1$ – the final temperature, $T_2$ – the initial temperature of the sample

Water and aluminium cup gain the heat, what is more water evaporates.

$Q_{\text{gained}} = c_{\text{Al}} m (T_1 - T_3) + c_{\text{w}} m_{\text{w}} (100 - T_3) + \lambda_{\text{w}} m_{\text{w}}$, where $T_3$ – the initial temperature of water and the cup, $\lambda_{\text{w}}$ – latent heat of vaporization of water.

$$c_{\text{al}} m (T_2 - T_1) = c_{\text{Al}} m (T_1 - T_3) + c_{\text{w}} m_{\text{w}} (100 - T_3) + \lambda_{\text{w}} m_{\text{w}}$$ $$c(\text{alloy}) = \frac{C_{\text{Al}} m (T_1 - T_3) + C_{\text{w}} m_{\text{w}} (100 - T_3) + \lambda_{\text{w}} m_{\text{w}}}{m (T_2 - T_1)}$$ $$c(\text{alloy}) = \frac{900 \times 0.2 \times (305 - 10) + 4200 \times 0.4 \times (100 - 10) + 2250000 \times 0.4}{0.15 \times (540 - 305)} = 31328 \, \text{J/kg/K}$$

Answer

The specific heat capacity of the alloy is 31328 J/kg/K

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