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 as 20°C. Scoffee = 4080 J/kg-K and Scup = 1020 J/kg-K.

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 as 20°C. Scoffee = 4080 J/kg-K and Scup = 1020 J/kg-K. (A) 45.5°C (B) 55.5°C (C) 65.5°C (D) 40.5°C

Tardigrade Admin · Accepted Answer

Let T be the final equal temperature.
Heat lost by coffee = Heat gained by cup

0.3 \times s_{co ff ee} \times (70 - T) = 0.12 \times s_{c u p} \times (T - 20)

\Rightarrow 0.3 \times 4080 (70 - T) = 0.12 \times 1020 \times (T - 20)

\Rightarrow 4 \times 70 - 4 T = 0.4 T - 8 \Rightarrow T = \frac{288}{4.4}

= 65. 5^{\circ} C

Q. 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 $_{co ff ee} = 4080 J / k g - K$ and $S_{c u p}$ = 1020 J/kg-K.

45.5 $^{\circ}$ C

55.5 $^{\circ}$ C

65.5 $^{\circ}$ C

40.5 $^{\circ}$ C

None of these

Q. 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 as 20∘C. Scoffee​=4080J/kg−K and Scup​ = 1020 J/kg-K.

45.5∘C

55.5∘C

65.5∘C

40.5∘C