Question

A 500-g aluminum container holds 300 g of water. The water and aluminum are initially at 40∘C. A 200-g iron block at 0∘C is added to the water. Assume the specific heat of iron is 450 J/kg⋅∘C, the specific heat of water 4180 J/kg⋅∘C and the specific heat of aluminum is 900 J/kg⋅∘C

.

1Determine the final equilibrium temperature. 2.Determine the change in thermal energy of the aluminum

3.Determine the change in thermal energy of the water. 4. Determine the change in thermal energy of the iron.

show work please

Answer #1

1.

m_{a} = mass of aluminium = 500 g = 0.5 kg

m_{w} = mass of water = 300 g = 0.3 kg

m_{i} = mass of iron = 200 g = 0.2 kg

T_{awi} = initial temperature of aluminium and water =
40

T_{ii} = initial temperature of iron = 0

T = final equilibrium temperature

c_{i} = specific heat of iron = 450

c_{w} = specific heat of water = 4180

c_{a} = specific heat of aluminium = 900

using conservation of heat

Heat lost by aluminium and water = Heat gained by iron

m_{w} c_{w} (T_{awi} - T) +
m_{a} c_{a} (T_{awi} - T) = m_{i}
c_{i} (T - T_{ii})

(0.3) (4180) (40 - T) + (0.5) (900) (40 - T) = (0.2) (450) (T - 0)

T = 38 C

2.

Q_{a} =
change in thermal energy of aluminium = m_{a} c_{a}
(T_{awi} - T) = (0.5) (900) (40 - 38) = 900 J

3.

Q_{w} =
change in thermal energy of water = m_{w} c_{w}
(T_{awi} - T) = (0.3) (4180) (40 - 38) = 2508 J

4.

Q_{i} =
change in thermal energy of iron = m_{i} c_{i} (T -
T_{ii} ) = (0.2) (450) (38 - 0) = 3420 J

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