Question

45 g of water at 50°C in a beaker is cooled when 50 g of copper at 18° C is added to it. The contents are stirred till a final constant temperature is reached. Calculate this final temperature. The specific heat capacity of copper is 0.39 J g^-1K^-1 and that of water is 4.2 J g^-1K^-1. State the assumption used.

Answer 1

Mass of water (m₁) = 45 g
Temperature of water (T₁) = 50°C
Mass of copper (m₂) = 50 g
temperature of copper (T₂) = 18°C
Final temperature (T) = ?

The specific heat capacity of the copper c₂ = 0.39 J/g/K
The specific heat capacity of water c₁ = 4.2 J/g/K

m₁c₁ (T₁ - T) = m₂ c₂ (T - T₂) 

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

`T = (45 xx 4.2 xx 50 + 50 xx 0. 39 xx 18)/ (45 xx 4.2 + 50 xx 0.39)`

= `9801 / 208.5`

 T = 47°C