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Question
An aluminium vessel of mass 0.5 kg contains 0.2 kg of water at 20°C. A block of iron of mass 0.2 kg at 100°C is gently put into the water. Find the equilibrium temperature of the mixture. Specific heat capacities of aluminium, iron and water are 910 J kg−1 K−1, 470 J kg−1 K−1 and 4200 J kg−1 K−1 respectively.
Solution
Given:-
Mass of aluminium = 0.5 kg
Mass of water = 0.2 kg
Mass of iron = 0.2 kg
Specific heat of aluminium = 910 J kg−1 K−1
Specific heat of iron = 470 J kg−1 K−1
Specific heat of water = 4200 J kg−1 K−1
Let the equilibrium temperature of the mixture be T.
Temperature of aluminium and water = 20°C = 273+20 = 293 K
Temperature of iron = 100°C = 273 + 100 = 373 K
Heat lost by iron, H1 = 0.2 × 470 × (373 − T)
Heat gained by water = 0.2 × 4200 × (T − 293)
Heat gained by aluminium = 0.5 × 910 × (T−293)
Total heat gained by water and iron, H2 = 0.5 × 910 (T−293) + 0.2 × 4200 × (T − 293)
H2 = (T − 293) [0.5 × 910 + 0.2 × 4200]
We know,
Heat gain = Heat lost
⇒ (T − 293) [0.5 × 910 + 0.2 × 4200] = 0.2 × 470 × (373 −T)
⇒ (T − 293) (455 + 840) = 94 (373 − T)
`rArr(T-293)1295/94=(373-T)`
`(T-293)xx14=(373-T)`
⇒ 14 T − 293 × 14 = 373 − T
⇒ 15 T = 373 + 4102 = 4475
`rArrT=4475/15=298.33Kapprox298K`
∴ T = (298 − 273)°C = 25°C
∴ Final temperature = 25°C
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