Question

# What is the total heat flow if 28 grams of water at 12°C is cooled to...

What is the total heat flow if 28 grams of water at 12°C is cooled to become ice at –19°C? The specific heat of liquid water is 4.18 J/g • °C; the specific heat of ice is 2.1 J/g • °C. The heat of fusion of ice is 333 J/g, and the freezing point of water is 0.0°C.

#### Homework Answers

Answer #1

Energy required to remove energy from 12°C to 0°C

Q1 = mass * specific heat * ∆T

= 28*4.18*(12-0)

= 1404.48 joules

Energy removed to change the phase

Q2 = mass * ∆H(fus)

= 28*333

= 9324 joules

Energy remived to change 0°C to -19°C

Q3 = mass * specific heat of ice * ∆T

= 28*2.1*(0-(-19)

= 1117.2 joules

Total heat = Q1 + Q2 + Q3

= 1404.48 + 9324 + 1117.2

= 11845.68 joules

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