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

thanks :)

Secure Information

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?

Sign Up