What is the final temperature when 2.50 x 10^5 J are added to 0.950 kg of ice at 0.0 degrees C? The specific heat capacity of water is 4186 J/(kg·degrees C) and the latent heat of fusion for ice is 3.35×10^3 J/kg.
2 answers
Please use steps!! Thanks
How much heat is required to melt the ice at 0C? That's
0.950 kg x 3.35E3 J/kg = ?
Subtract that from 2.50E5 J to determine the excess heat available to heat the 0.950 kg water at 0C left. Then
q(what's left) = mass x specific heat liquid H2O x (Tfinal-Tinitial) and solve for Tfinal.
The answer is approx 60 C.
0.950 kg x 3.35E3 J/kg = ?
Subtract that from 2.50E5 J to determine the excess heat available to heat the 0.950 kg water at 0C left. Then
q(what's left) = mass x specific heat liquid H2O x (Tfinal-Tinitial) and solve for Tfinal.
The answer is approx 60 C.
7 answers