Answer to Question #288288 in Molecular Physics | Thermodynamics for harry

Question

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?

Answer 1

$Q=cm\Delta{T}=3.67\frac{kJ}{kg\cdot\degree{C}}\times0.750\ kg\times(80\degree{C-25\degree{C}})=151\ kJ$

Answer: 151 kJ