Question

50 g of a hot solid of specific heat capacity 0.25 Jg⁻¹°C⁻¹ and at 100°C is placed in 80 g of cold water, when the temperature of cold water rises by 3°C. Find the initial temperature of the cold water.

Answer 1

Let the initial temperature of the cold water be x°C.

Since cold water rises by 3°C, final temperature of the mixture θ = (x + 3)°C

Heat lost by the hot solid = m_sc_s (100 − θ)

= 50 × 0.25 × [100 − (x + 3)]

= 12.5(97 − x) J

Heat gained by cold water = m_wc_w × 3

= 80 × 4.2 × 3

= 1008 J

By the principle of calorimetry,

12.5(97 − x) = 1008

97 − x = 80.64

x = 16.36°C

Thus, the initial temperature of the cold water is 16.36°C.