Question

200 g of hot water at 80°C is added to 400 g of cold water at 10°C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = 4200 J kg^-1K^-1.

Answer 1

Mass of hot water (m₁) = 200g

Temperature of hot water (T₁) = 80°C

Mass of cold water (m₂) = 400g

Temperature of cold water (T₂) =10°C

Final temperature (T) = ?

m₁ c₁ (T₁ - T) = m₂ c₂ (T - T₂)

c₁ = c₂

T = `("m"_1 "T"_1  + "m"_2 "T"_2)/("m"_2 + "m"_1)`

`T = (200 xx 80 + 400 xx 10)/600`

T = 33.3°C