Question

10 kg of water at 90°C is cooled to 60°C. Calculate the heat lost by the water. Specific heat of water is 4200 J kg1 °C1.

Answer 1

Answer:

1260 kilojoules

Explanation:

Mass=10kg

Specific heat= 4200J kg^-1 °C^-1

Heat lost= 90°C-60°C = 30°C

Heat lost = 10kg×4200J÷kg×°C×30°C

= 10×4200J×30

= 12,60,000J

1000joule=1kilojoule

= 12,60,000J= 12,60,000÷1000 kilojoule

= 1260 kilojoules = 1260 kJ