SME
Grade 11 Science · Problem type
Melting and boiling: latent heat
Use this for “How much heat turns 0.5 kg of ice at 0 °C into steam at 100 °C, and how long does each stage take with a 500 W heater?”
Chart
Melting point = 0 °C
Boiling point = 100 °C
Time = 0 + 334 + 418 + 2,260 = 3,012 s
Mass of ice: 0.5 kg
Heater power: 500 W
Heat to melt it: 167 kJ
Heat to warm the water to 100 °C: 209 kJ
Heat to boil it away: 1,130 kJ
Heat from ice to steam: 1,506 kJ
This is an example. Type your own values to start a new problem.
m
Mass of ice
example · kg
P
Heater power
example
Qf
Heat to melt it
calculated · kJ
tf
Time to melt
calculated
Qw
Heat to warm the water to 100 °C
calculated · kJ
tw
Time to warm the water
calculated
Qᵥ
Heat to boil it away
calculated · kJ
tᵥ
Time to boil
calculated
Qₜₒₜₐₗ
Heat from ice to steam
calculated · kJ
Formulas
Qf
=
m
×
334
167 = 0.5 × 334
tf
=
1,000
×
Qf/
P
334
=
1,000
×
167
500
Qw
=
m
×
4.18
×
100
209 = 0.5 × 4.18 × 100
tw
=
1,000
×
Qw/
P
418
=
1,000
×
209
500
Qᵥ
=
m
×
2,260
1,130 = 0.5 × 2,260
tᵥ
=
1,000
×
Qᵥ/
P
2,260
=
1,000
×
1,130
500
Qₜₒₜₐₗ = Qf + Qw + Qᵥ
1,506 = 167 + 209 + 1,130
Assumptions
•
While ice melts or water boils the temperature stays put: the heat breaks bonds, Q = mL.
•
Ice melts with Lf = 334 kJ/kg; water boils away with Lᵥ = 2260 kJ/kg; warming water takes 4.18 kJ/(kg·°C).
•
The heater gives P joules each second, all of it to the water: t = Q ÷ P.
Step-by-step
WE KNOW
m = 0.5 kg, P = 500 W
FIND
heat to melt it (Qf), time to melt (tf), heat to warm the water to 100 °C (Qw), time to warm the water (tw), heat to boil it away (Qᵥ), time to boil (tᵥ), heat from ice to steam (Qₜₒₜₐₗ)
Step 1 · Find heat to melt it (Qf)
Use Qf = m × 334
Each kilogram of ice takes 334 kJ to melt.
Qf = 0.5 × 334
Qf = 167 kJ
Step 2 · Find time to melt (tf)
Use tf = 1,000 × Qf/P
A watt is a joule each second: × 1000 turns the kJ into joules, then divide by the power.
tf
=
1,000
×
167
500
tf
=
167,000
500
tf = 334 s
Step 3 · Find heat to warm the water to 100 °C (Qw)
Use Qw = m × 4.18 × 100
Warming the melted water from 0 °C to 100 °C: mcΔT with c = 4.18 kJ/(kg·°C).
Qw = 0.5 × 4.18 × 100
Qw = 2.09 × 100
Qw = 209 kJ
Step 4 · Find time to warm the water (tw)
Use tw = 1,000 × Qw/P
× 1000 turns the kJ into joules, then divide by the power.
tw
=
1,000
×
209
500
tw
=
209,000
500
tw = 418 s
Step 5 · Find heat to boil it away (Qᵥ)
Use Qᵥ = m × 2,260
Each kilogram of water takes 2260 kJ to boil away.
Qᵥ = 0.5 × 2,260
Qᵥ = 1,130 kJ
Step 6 · Find time to boil (tᵥ)
Use tᵥ = 1,000 × Qᵥ/P
× 1000 turns the kJ into joules, then divide by the power: boiling takes much longer than melting.
tᵥ
=
1,000
×
1,130
500
tᵥ
=
1,130,000
500
tᵥ = 2,260 s
Step 7 · Find heat from ice to steam (Qₜₒₜₐₗ)
Use Qₜₒₜₐₗ = Qf + Qw + Qᵥ
Add the heat for each stage: melt, warm, boil.
Qₜₒₜₐₗ = 167 + 209 + 1,130
Qₜₒₜₐₗ = 1,506 kJ
Check
Put the numbers back in. Both sides should match.
167 = 0.5 × 334 ✓
334
=
1,000
×
167
500
✓
209 = 0.5 × 4.18 × 100 ✓
418
=
1,000
×
209
500
✓
1,130 = 0.5 × 2,260 ✓
2,260
=
1,000
×
1,130
500
✓
1,506 = 167 + 209 + 1,130 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Refresh
Related lessons
Thermal energy, heat and the laws of thermodynamics
Main lesson
Heat to warm something: Q = mcΔT
Use this for “How much heat warms 2 kg of water from 15 °C to 65 °C?”
The first law of thermodynamics
Use this for “A gas takes in 500 J of heat and does 200 J of work pushing a piston out. How much does its internal energy change?”
A heat engine and the Carnot limit
Use this for “An engine takes in 2000 J from a 600 K source and does 500 J of work. How much heat goes to the 300 K sink, and how efficient is it compared with the most it could be?”
A refrigerator
Use this for “A fridge removes 900 J from its 270 K inside using 300 J of work. How much heat reaches the 300 K kitchen, and what are its COP and the best possible COP?”
Which law of thermodynamics?
Use this for “Which law of thermodynamics explains this?”