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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)

Respuesta :

joseaaronlara

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

                         

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