Calorimetry numerical **Question: 9** [reference: Selina ICSE class 10 Physics]

**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.**

**Solution:**

Say, the specific heat capacity of the metal = s

Heat liberated by metal = m × s × △t = 50 × s × (150 – 20)

Heat absorbed by water = m_{w} × s_{w} × △t = 100 × 4.2 × (20 – 11)

Heat energy lost = heat energy gained

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

s = 0.582 J g-1 K-1