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.

1. 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.

2. Determine the rate at which heat is removed from 1.5 kg of water at 20°C to make ice at -12°C by a refrigerator if it is kept on for 2 hours.

Answer 1

Answer on Question #80268, Physics / Other

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.

1. 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 30.5°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 sum of the internal energy changes of alloy sample (1), water (2) and aluminium cup(3) equals zero:

$$m_1 \cdot C_1 \cdot (T_{i1} - T_f) + m_2 \cdot C_2 \cdot (T_{i2} - T_f) + m_3 \cdot C_3 \cdot (T_{i3} - T_f) = 0$$

So,

$$C_1 = \frac{(m_2 \cdot C_2 + m_3 \cdot C_3) \cdot (T_f - T_{i2})}{m_1 \cdot (T_{i1} - T_f)} = \frac{(0.4 \cdot 4200 + 0.2 \cdot 900) \cdot (30.5 - 10)}{0.15 \cdot (540 - 30.5)} = 498.9 \frac{J}{kg\,K} \approx 500 \frac{J}{kg\,K}$$

Answer: $500\frac{J}{kg\,K}$

2. Determine the rate at which heat is removed from 1.5 kg of water at 20°C to make ice at -12°C by a refrigerator if it is kept on for 2 hours.

Solution:

Total heat removed from water to make ice is

$$Q = Q_1 + Q_2 + Q_3 = C_{water} m_{water} (\Delta T_1) + L_{ice} m_{water} + C_{ice} m_{water} (\Delta T_2) = \\ = m_{water} (20 \cdot C_{water} + L_{ice} + 12 \cdot C_{ice}) = \\ = 1.5 \cdot (20 \cdot 4200 + 334000 + 12 \cdot 2100) = 664800\,J$$

The rate is

$$\dot {Q} = \frac {Q}{t} = \frac {664800 \, J}{2 \, hr \cdot 3600 \, s} = 92.3 \, J/s$$

Answer: $92.3 \, J/s$

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