Answer:
The answer to your question is Qt = 15750 cal
Explanation:
Data
mass = m = 150 g
Initial temperature = T1 = 0°C
Final temperature = T2 = 25°C
Process
1.- Calculate the Latent heat of fusion
Latent heat of fusion for water = l = 80 cal/g
Heat = Q
Formula
Q = ml
Substitution
Q = (150)(80)
Simplification and result
Q = 12000 cal
2.- Calculate the Specific heat
Specific heat of water = C = 1 cal/g°C
Formula
Q = mCΔT
Substitution
Q = (150)(1)(25 - 0)
Simplification and result
Q = 3750 cal
3.- Calculate the total heat
Qt = 12000 + 3750
Qt = 15750 cal
