Heat and change of phase – problems and solutions

1. Based on the graph, what is the heat absorbed by 5-kg water during process C-D ? Specific heat of water = 4,200 J/kg oC.

Known :Heat and change of phase – problems and solutions 1

Mass (m) = 5 kg

Specific heat of water (c) = 4200 J/kg oC

Initial temperature (T1) = 0 oC

Final temperature (T2) = 10 oC

The change in temperature (ΔT) = 10 oC – 0 oC = 10 oC

Wanted : Heat (Q)

Solution :

Q = m c ΔT = (5 kg)(4200 J/kg oC)(10 oC) = (5)(4200 J)(10) = (50)(4200 J) = 210,000 Joule

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2. What is the heat absorbed by 20 gram ice during process A-C, if the specific heat of ice is 2100 J/kg oC and heat of fusion for ice is 336,000 J/kg.

Heat and change of phase – problems and solutions 2Known :

Mass of ice (m) = 200 gram = 200/1000 kilogram = 0.2 kilogram

Specific heat (c) = 2100 J/kg oC

Heat of fusion for ice (LF) = 336,000 J/kg

Wanted: Heat absorbed by ice during process A-C

Solution :

During process A-B, heat is absorbed by ice to change its temperature from -10oC to 0oC.

Q = m c ΔT

Q = heat, m = mass of ice, c = specific heat of ice, ΔT = the change in temperature

Q1 = m c ΔT = (0.2)(2100)(0-(-10) = (0.2)(2100)(10) = (2)(2100) = 4200 Joule

During process B-C, heat is absorbed to melting all ice become water. In this process only changes in phase and there is no change in temperature. The ice and water temperature remains 0oC.

Heat absorbed during process B-C :

Q2 = m LF = (0.2 kg)(336,000 J/kg) = 67,200 Joule

Heat absorbed during process A-C :

Q = Q1 + Q2 = 4200 Joule + 67,200 Joule = 71,400 Joule

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3. A 2-kg water is heated, and heat of vaporization for water = 2.27 x 106 J/kg, specific heat for water = 4,200 J/kg oC and atmosphere pressure is 1 atm, what is the heat absorbed by water during process B-C ?

Known :Heat and change of phase – problems and solutions 3

Mass (m) = 2 kg

Heat of vaporization for water (LV) = 2.27 x 106 J/kg

Specific heat for water (c) = 4,200 J/kg oC

Wanted : Heat absorbed during process B-C.

Solution :

In process A-B, heat absorbed to increases the temperature of water from 60oC to 100oC.

Q = m LV = (2 kg)(2.27 x 106 J/kg) = 4.54 x 106 Joule = 4540 x 103 Joule = 4540 kilo Joule

4. Based on figure below, if mass of ice is 500 gram, specific heat is 2100 J/kg oC and latent heat of fusion of ice is 336,000 J/kg. Determine heat required in process P–Q–R.

Known :Heat and change of phase – problems and solutions 5

Mass of ice (m) = 500 gram = 500/1000 kg = 0.5 kg

Specific heat for ice (c) = 2100 J/kg oC

Latent heat of fusion (LF) = 336,000 J/kg

Wanted : Heat

Solution :

Process P-Q = heat absorbed to raise the ice temperature from -4oC to 0oC

Process Q-R = heat absorbed by ice to melting all the ice into water

Heat absorbed during process P-Q calculated using equation :

Q = m c ΔT

Q = heat, m = mass, c = specific heat of ice, ΔT = the change in temperature

Heat absorbed ice during process P-Q calculated using this equation :

Q = (0.5 kg)(2100 J/kg oC)(0oC-(-4oC))

Q = (0.5 kg)(2100 J/kg oC)(0oC + 4oC)

Q = (0.5 kg)(2100 J/kg oC)(4oC)

Q = (0.5)(2100 J)(4)

Q = 4200 Joule

Heat absorbed during process Q-R calculated using this equation :

Q = m LF

Q = (0.5 kg)(336,000 J/kg)

Q = (0.5)(336,000 J)

Q = 168,000 Joule

Total heat :

Q = 4200 Joule + 168,000 Joule

Q = 172,200 Joule

Q = 172.2 kilo Joule

Q = 172.2 kJ

5. Ice with mass of 50 gram heated from -5oC into water at 60oC. If heat of fusion for ice is 80 cal/gram, specific heat for ice is 0.5 cal/gram oC, specific heat for water is 1 cal/gram oC, determine heat required in process C-D.

Known :

Mass of ice = mass of water (m) = 50 gram Heat and change of phase – problems and solutions 6

Latent heat of fusion for ice (LF) = 80 cals/gram

Specific heat for ice (c ice) = 0.5 cal/gram oC

Specific heat for water (c water) = 1 cal/gram oC

Wanted: Heat required in process C-D

Solution :

In process A to B, heat used to raise the temperature of ice from -5oC to 0oC. Heat calculated using this equation : Q = m c ΔT, where Q = heat, m = mass of ice, c = specific heat for ice, ΔT = the change in temperature from -5oC to 0oC.

In process B to C, heat used to melting all ice to water. In this process, temperature not changed. Heat calculated using this equation: Q = m LF, where Q = heat, m = mass of ice, LF = latent heat of fusion of ice.

In process C to D, heat used to raise the temperature of water from 0oC to 60oC. Heat calculated using this equation: Q = m c ΔT, where Q = heat, m = mass of water, c = specific heat of water, ΔT = the change in temperature from 0oC to 60oC.

The heat required from process C to D :

Q = m c ΔT = (50 gram)(1 cal/gram oC)(60oC-0oC) = (50)(1)(60) cal = 3000 cal

6. If the mass of ice is 1 kg, the specific heat of ice is 2100 J/kgoC, latent heat of fusion for ice is 336,000 J/kg and specific heat of water is 4200 J/kgoC, then determine heat required in process P-Q-R.

Known :

Mass of ice = 1 kgHeat and change of phase – problems and solutions 7

Latent heat of fusion of ice (LF) = 336,000 J/kg

Specific heat of ice (c ice) = 2,100 J/kgoC

Specific heat of water (c water) = 4,200 J/kgoC

Wanted: Heat required in process P-Q-R

Solution :

In process P to Q, heat used to raise the temperature of ice from -5oC to 0oC. Heat calculated using this equation : Q = m c ΔT, where Q = heat, m = mass of ice, c = specific heat of ice, ΔT = the change in temperature from -5oC to 0oC.

Heat required in process P to Q :

Q = m c ΔT = (1)(2100)(0-(-5)) = (2100)(5) = 10500 Joule

In process Q to R, the heat required to melting all ice into water. In this process, temperature is unchanged. Heat calculated using this equation: Q = m LF, where Q = heat, m = mass of ice, LF = latent heat of fusion of ice.

Heat required in process Q to R :

Q = m LF = (1)(336,000) = 336,000 Joule

Total heat :

10,500 + 336,000 = 346,500 Joule