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:

According to the statement, mass of the metal is 50g and the change in temperature is equal to (150 – 20)

Then using the expression for heat energy, we have –

Heat lost by metal = m × s × △t

Heat lost by metal = 50 × s × (150 – 20)

Similarly, the mass of water is 100g and the change in temperature is equal to (20 – 11)

And the specific heat capacity of water is 4.2 J

then, the heat gained by water = m_w × s_w × △t

Heat gained by water = 100 × 4.2 × (20 – 11)

Since, the Heat energy lost by metal is equal to the heat energy gained by water, we have

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

Upon solving, we get

s = 0.582 J g^-1 K^-1