Question

0.3 kg of hot coffee, which is at 70$^{\circ}$C, is poured into a cup of mass 0.12 kg. Find the final equilibrium temperature. Take room temperature as 20$^{\circ}$C.
S$_{coffee} = 4080 J/kg-K $ and $ S_{cup}$ = 1020 J/kg-K.

• A. 45.5$^{\circ}$C
• B. 55.5$^{\circ}$C
• C. 65.5$^{\circ}$C
• D. 40.5$^{\circ}$C
• E. None of these

Answer 1

Answer: (C)

Let T be the final equal temperature.
Heat lost by coffee = Heat gained by cup
$0.3 \times s_{coffee} \times (70-T)=0.12 \times s_{cup} \times (T-20)$
$\Rightarrow \, \, \, \, 0.3 \times 4080(70-T) =0.12 \times 1020 \times (T-20)$
$\Rightarrow \, \, \, \, 4 \times 70 -4T =0.4 T -8 \Rightarrow \, \, T=\frac{288}{4.4}$
$ \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, =65.5 ^{\circ}C$