Question

200 g of hot water at 80°C is added to 300 g of cold water at 10°C. Calculate the final temperature of the mixture of water. Consider the heat taken by the container to be negligible. [Specific heat capacity of water is 4200 J kg⁻¹ °C⁻¹]

Answer 1

Given: Mass of hot water (m₁) = 200 g

Initial temperature of hot water (T₁) = 80°C

Mass of cold water (m₂) = 300 g

Initial temperature of cold water(T₂) = 10°C

Specific heat capacity of water (c) = 4200 J kg⁻¹ °C⁻¹

Let the final temperature of the mixture = T_f

By the principle of the calorimeter:

Heat given = Heat taken

m₁ × c × (T₁ − T_f) = m₂ × c × (T_f − T₂)

200 × c × (80 − T_f) = 300 × c × (T_f −10)

200 × 80 − 200T_f = 300T_f − 300 × 10

16000 − 200T_f = 300T_f − 3000

16000 + 3000 = 200T_f + 300 T_f

19000 = 500T_f

T_f = `19000/500`

T_f = 38°C