How much total energy must be absorbed by a 150 g sample of ice at 0.0°C that it melts

AND then warms to 25.0°C? (Break it into steps and add them together, phase change

and temperature change)

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