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}) = 0 ^{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, specific heat is 2100 J/kg ^{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 1 cal/gram ^{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

**What is latent heat?****Answer**: Latent heat is the heat energy required to change the phase of a substance without changing its temperature. It represents the energy needed to break or form intermolecular bonds during phase changes.

**Why does the temperature remain constant during a phase change?****Answer**: During a phase change, all the heat energy provided or removed is used to change the substance’s phase. This energy is used to break or form intermolecular bonds, not to increase the substance’s kinetic energy or temperature.

**What is the difference between latent heat of fusion and latent heat of vaporization?****Answer**: The latent heat of fusion refers to the heat energy required to change a substance from solid to liquid (or vice versa) at its melting point. The latent heat of vaporization refers to the heat energy required to change a substance from liquid to gas (or vice versa) at its boiling point.

**Why do we feel cold when we get out of a swimming pool on a windy day, even if the air temperature is warm?****Answer**: As water evaporates from our skin, it undergoes a phase change from liquid to vapor. This process requires energy, which is taken from our skin in the form of heat, causing our skin to feel cooler.

**Why is steam at 100°C more dangerous than boiling water at 100°C?****Answer**: Steam contains more energy than boiling water because it has absorbed the latent heat of vaporization during the phase change from water to steam. When steam condenses on the skin, it releases this extra energy, causing more severe burns than boiling water.

**How does adding salt to ice lower its melting point?****Answer**: Adding salt disrupts the equilibrium between the melting and freezing of the ice-water mixture. This causes more ice to melt at a lower temperature than usual, effectively lowering the freezing point of the mixture.

**Why does sweating help cool the body on a hot day?****Answer**: Sweat, when evaporated from the skin, undergoes a phase change from liquid to gas. This evaporation process requires energy, which is taken from the skin, thus cooling it.

**What is sublimation in the context of phase changes?****Answer**: Sublimation is the direct phase change from a solid to a gas without passing through the liquid phase. A common example is dry ice (solid carbon dioxide) transitioning directly into carbon dioxide gas.

**Why is heat energy required to melt ice even though the temperature remains at 0°C?****Answer**: The heat energy is used to break the intermolecular bonds holding the ice in its solid structure. Once these bonds are broken, the ice transitions to the liquid phase, even though there’s no change in temperature.

**How does pressure affect the boiling point of a liquid?**

**Answer**: Generally, increasing the external pressure raises the boiling point of a liquid, while decreasing the pressure lowers the boiling point. This is why water boils at temperatures lower than 100°C at high altitudes, where atmospheric pressure is lower.