Question

`0.3 Kg` of hot coffee, which is at `70^(@)C`, is poured into a cup of mass 0.12 kg . Find the final equilibrium temperature. Take room temperature at `20^(@)C` .
(`s_(coffee) = 4080 J/kg-K and s_(cup) = 1020 J/kg-K`.)
A. `45.5^(@)C`
B. `55.5^(@)C`
C. `65.5^(@)C`
D. `40.5^(@)C`

Answer 1

Correct Answer - C
Let T be the final equilibrium temperature.
Heat lost by coffee= Heat gained by cup
`0.3 xx c_(coffee) xx (70-T) = 0.12 xx c_(Cup) xx (T-20)`
`rArr (0.3xx 4080(70-T) = 0.12 xx 1020 xx (T-20)`
`rAr 4 xx 70 - 4T = 0.4T-8` `rArrT=(288/4.4) = 65.5^(@)`C