the heat capacity of ice is 2.108 kJ/ (kg*K) the latent heat of ice is 334.7 kJ/kg,
let the final temperature of the water is T,
then 390kJ - 0.9*2.108 *5-0.9*334.7 = 79.3 kJ
T = 79.3KJ/(0.9*4.184) = 21 C
can u answer part b?
yes, it need to increase by a factor 3.6 too
Content will be erased after question is completed.
Enter the email address associated with your account, and we will email you a link to reset your password.
Forgot your password?