Question

14. How much heat does it take to boil (100°C) 3.50 kg of water from room temperature (22°C)? (cwater = 4180 J/kg C)

Answer 1

Solution.

$m=3.50kg;$

$t_1=22^oC;$

$t_2=100^oC;$

$c=4180J/kgC;$

$Q-?;$

$Q=cm\Delta t;$

$Q=4180J/kgC\sdot3.50kg\sdot\sdot(100-22)^oC=14552J;$

Answer: $Q=14552J.$