Question

600 g of copper at 50°C is mixed with lOOOg water at 20°C. Find the final temperature of the mixture. The specific heat capacity of copper is 0.4 Jg^-1°C^-1 and that of water is 4.2 Jg^-1°C^-1

Answer 1

Let final temperature of the mixture of copper and water = x °C
For copper:
Mass of copper m₁ = 600g
Specific heat capacity of copper c₁ = 0.4Jg^-1°C^-1
Initial temperature of copper t₁ = 50°C
Final temperature of copper t₂ = x°C
Fall in temperature Δt = (50 – x)°C
Heat lost by copper = m₁ × c₁ × t
= 600 × 0.4 × (50 – x)

For water:
Mass of water m₂ = 1000g
Specific heat capacity of water c₂ = 4.2Jg^-1°C^-14
Initial temperature of water t₁ = 20°C
Final temperature of water t₂ = x°C
Rise in temperature Δt = (x – 20)°C
Heat gained by water = m₂ × c₂ × t
= 100 × 4.2 × (x – 20)

According to the principle of calorimetry,
Heat lost by copper = Heat gained by water
600 × 0.4 × (50 – x) =1000 x 4.2 x (x – 20)
240 × (50 – x) = 4200 ( x- 20)
12,000 – 240x = 4200x-84000
4200x + 240x =12000 + 84000
4440 = 96000
x = \(\frac{96000}{4440}=21.6\)
So, the final temperature of the mixture of water and copper x = 21.6°C