Answer to Question #161363 in Molecular Physics | Thermodynamics for Martinez

Question

 Calculate the quantity of heat absorbed by 3 kg of copper when heated from room temperature to 70oC. (ccu= 0.092 cal/groC)

Answer 1

$$Q=mc\Delta T,$$ $$Q=3000\ g\cdot0.092\ \dfrac{cal}{g\cdot \!^{\circ}C}\cdot(70\ ^{\circ}C-20\ ^{\circ}C)=13800\ cal=13.8\ kcal.$$