# 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 ^{o}C.

__Known :__

Mass (m) = 5 kg

Specific heat of water (c) = 4200 J/kg ^{o}C

Initial temperature (T_{1 }

^{o}C

Final temperature (T_{2}) = 10 ^{o}C

The change in temperature (ΔT) = 10 ^{o}C – 0 ^{o}C = 10 ^{o}C

__Wanted :__ Heat (Q)

__Solution :__

Q = m c ΔT = (5 kg)(4200 J/kg ^{o}C)(10 ^{o}C) = (5)(4200 J)(10) = (50)(4200 J) = 210,000 Joule

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

__Known :__

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

Specific heat (c) = 2100 J/kg ^{o}C

Heat of fusion for ice (L_{F}) = 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 -10^{o}C to 0^{o}C.

Q = m c ΔT

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

Q_{1 }= 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 0^{o}C.

Heat absorbed during process B-C :

Q_{2 }= m L_{F} = (0.2 kg)(336,000 J/kg) = 67,200 Joule

Heat absorbed during process A-C :

Q = Q_{1 }+ Q_{2 }= 4200 Joule + 67,200 Joule = 71,400 Joule

3. A 2-kg water is heated, and heat of vaporization for water = 2.27 x 10^{6} J/kg, specific heat for water = 4,200 J/kg ^{o}C and atmosphere pressure is 1 atm, what is the heat absorbed by water during process B-C ?

__Known :__

Mass (m) = 2 kg

Heat of vaporization for water (L_{V}) = 2.27 x 10^{6} J/kg

Specific heat for water (c) = 4,200 J/kg ^{o}C

__Wanted :__ Heat absorbed during process B-C.

__Solution :__

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

^{o}C to 100

^{o}C.

Q = m L_{V} = (2 kg)(2.27 x 10^{6} J/kg) = 4.54 x 10^{6} Joule = 4540 x 10^{3 }Joule = 4540 kilo Joule

4. Based on figure below, if mass of ice is 500 gram, specifi c heat is

^{o}C and latent heat of fusion of ice is 336,000 J/kg. Determine heat required in process P–Q–R.

__Known :__

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

Specific heat for ice (c) = 2100 J/kg ^{o}C

Latent heat of fusion (L_{F}) = 336,000 J/kg

__Wanted :__ Heat

__Solution :__

Process P-Q = heat absorbed to raise the ice temperature from -4^{o}C to 0^{o}C

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 ^{o}C)(0^{o}C-(-4^{o}C))

Q = (0.5 kg)(2100 J/kg ^{o}C)(0^{o}C + 4^{o}C)

Q = (0.5 kg)(2100 J/kg ^{o}C)(4^{o}C)

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

Q = 4200 Joule

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

Q = m L_{F}

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 -5^{o}C into water at 60^{o}C. If heat of fusion for ice is 80 cal/gram, specific heat for ice is 0.5 cal/gram ^{o}C, specific heat for water is

^{o}C, determine heat required in process C-D.

__Known :__

Mass of ice = mass of water (m) = 50 gram

Latent heat of fusion for ice (L_{F}) = 80 cals/gram

Specific heat for ice (c ice) = 0.5 cal/gram ^{o}C

Specific heat for water (c water) = 1 cal/gram ^{o}C

__Wanted:__ Heat required in process C-D

__Solution :__

In process A to B, heat used to raise the temperature of ice from -5^{o}C to 0^{o}C. 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 -5^{o}C to 0^{o}C.

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 L_{F}, where Q = heat, m = mass of ice, L_{F }= latent heat of fusion of ice.

In process C to D, heat used to raise the temperature of water from 0^{o}C to 60^{o}C. 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 0^{o}C to 60^{o}C.

The heat required from process C to D :

Q = m c ΔT = (50 gram)(1 cal/gram ^{o}C)(60^{o}C-0^{o}C) = (50)(1)(60) cal = 3000 cal

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

__Known :__

Mass of ice = 1 kg

Latent heat of fusion of ice (L_{F}) = 336,000 J/kg

Specific heat of ice (c ice) = 2,100 J/kg^{o}C

Specific heat of water (c water) = 4,200 J/kg^{o}C

__Wanted:__ Heat required in process P-Q-R

__Solution :__

In process P to Q, heat used to raise the temperature of ice from -5^{o}C to 0^{o}C. 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 **-5*^{o}*C **to **0*^{o}*C.*

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 L*_{F}*, **where **Q = **heat**, m = mass **of ice**, L*_{F }*= **latent heat of fusion of ice**.*

Heat required in process Q to R :

Q = m L_{F }= (1)(336,000) = 336,000 Joule

Total heat :

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