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.

Answer 1

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