Question

A mass of 50 g of a certain metal at 150° C is immersed in 100 g of water at 11° C. The final temperature is 20° C. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is 4.2 J g^-1 K^-1.

Answer 1

Heat liberated by metal = m × s × Δ t

= 50 × s × (150 – 20)

Heat absorbed by water = m_w × s_w × Δt

= 100 × 4.2 × (20 – 11)

Heat energy lost = heat energy gained

⇒ 50 × s × (150 – 20) = 100 × 4.2 × (20 – 11)

⇒ 50 × s × 130 = 100 × 4.2 × 9

⇒ s = `(100 × 4.2 × 9)/(50 xx 130)`

s = `37.8/65`

S = 0.58 J g^-1 K^-1.