A container has 2.4 liters of water at 20^C. What is the amount of heat required to boil the water? Write final answer to the nearest whole number and indicate the unit in kJ.
2.4l= 2400ml
density= 1ml/g
Mass= 1× 2400= 2400g= 2.4kg
Water boils at 0°C
Boiling point = 100°C
Initial = 20°C
Specific heat capacity of water = 4200j/kg/°C
= 4200×(100-20)×2.4
= 806400j
= 806.400kJ
Comments
Leave a comment