Question

An aluminium vessel of mass 0.5kg contains 0.2kg of water at 20°C. A block of iron of mass 0.2kg 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 910J/kg-K, 470J/kg-K and 4200J/kg-K respectively.

Answer 1

Mass of aluminium = 0.5kg 

Mass of iron = 0.2kg

Specific heat of iron = 100^∘C = 373^∘K

Specific heat of iron 470J Kg⁻¹K⁻¹

Mass of water = 0.2kg

Temp. of aluminium and water = 20^∘C = 297^∘K

Specific heat of aluminium = 910J Kg⁻¹K⁻¹

Specific heat of water 4200J Kg⁻¹K⁻¹

So the

Heat gain = 0.5 × 910(T − 293) + 0.2 × 4200 × (343 − T)

∴ (T − 292) (0.5 × 910 + 0.2 × 4200)

also

Heat lost = 0.2 × 470 × (373 − T)

We know that, Heat gain = Heat lost

So we substitute and get

∴ (T − 292) (0.5 × 910 + 0.2 × 4200) = 0.2 × 470 × (373 − T)

∴ (T − 293) (455 + 8400) = 49(373 − T)

Hence

∴ (T − 293) (\(\frac{1295}{94}\)) = (373 − T)

∴ (T − 293) × 14 = 49(373 − T)

So the temperature is ⇒ T = \(\frac{4474}{15}\) = 298K

∴ T = 298 − 273 = 25^∘ − 0^∘C

Final temperature = 25^∘C

Answer 2

Mass of aluminium = 0.5kg, 

Mass of water = 0.2kg
Mass of Iron = 0.2kg 

Temp. of aluminium and water = 20°C = 297°k

Sp heat o f Iron = 100°C = 373°k. 

Sp heat of aluminium = 910J/kg-k

Sp heat of Iron = 470J/kg-k

Heat again = 0.5 × 910(T – 293) + 0.2 × 4200 × (343 –T) 

= (T – 292) (0.5 × 910 + 0.2 × 4200) Heat lost = 0.2 × 470 × (373 – T) 

∴ Heat gain = Heat lost 

=> (T – 292) (0.5 × 910 + 0.2 × 4200) = 0.2 × 470 × (373 – T) 

=> (T – 293) (455 + 8400) = 49(373 – T) 

=> (T – 293)(1295/94)  = (373 – T) 

=> (T – 293) × 14 = 373 – T 

=> T = 4475/15 = 298 k 

=> T = 298 – 273 = 25°C. 

The final temp = 25°C.