750 g of hot curry at 80⁰C cools down to 25⁰C after 1 hour left in the kitchen. If the specific heat of the soup is 3.67kJkg∘C, how much energy does the curry release into the kitchen?
Q=cmΔT=3.67kJkg⋅°C×0.750 kg×(80°C−25°C)=151 kJQ=cm\Delta{T}=3.67\frac{kJ}{kg\cdot\degree{C}}\times0.750\ kg\times(80\degree{C-25\degree{C}})=151\ kJQ=cmΔT=3.67kg⋅°CkJ×0.750 kg×(80°C−25°C)=151 kJ
Answer: 151 kJ
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