In Calorimetry you used $Q = mc\Delta T$ to find the heat that warms an object. But heat a pot of ice and the thermometer stalls at $0\,^\circ\text{C}$, and again at $100\,^\circ\text{C}$, as you saw in Changes of State. Where does the heat go while the temperature stands still?
Why it works
Temperature measures the average kinetic energy of the molecules. During melting, the heat goes into pulling molecules out of the crystal's fixed places, raising their potential energy instead of their speed. So the temperature stays put until every bit of ice has melted. Boiling pulls the molecules completely apart, which takes far more energy, so $L_v$ is much bigger than $L_f$.
Key idea
Changing the state of a mass $m$ at its melting or boiling point takes $Q = mL$, with no change in temperature. For water, $L_f = 3.34 \times 10^{5}\,\text{J/kg}$ and $L_v = 2.26 \times 10^{6}\,\text{J/kg}$. For a whole heating job, add the heat of every stage.
| Stage | Formula | Constant for water |
|---|---|---|
| Warm ice | $Q = mc_{\text{ice}}\Delta T$ | $c_{\text{ice}} \approx 2100\,\text{J/(kg}\cdot\text{K)}$ |
| Melt at $0\,^\circ\text{C}$ | $Q = mL_f$ | $L_f = 3.34 \times 10^{5}\,\text{J/kg}$ |
| Warm water | $Q = mc_{\text{water}}\Delta T$ | $c_{\text{water}} = 4186\,\text{J/(kg}\cdot\text{K)}$ |
| Boil at $100\,^\circ\text{C}$ | $Q = mL_v$ | $L_v = 2.26 \times 10^{6}\,\text{J/kg}$ |
Worked examples
Worked example: Melting
How much heat melts $0.50\,\text{kg}$ of ice that is already at $0\,^\circ\text{C}$?
- At the melting point all the heat goes into the change of state: $Q = mL_f$.
- $Q = (0.50\,\text{kg})(3.34 \times 10^{5}\,\text{J/kg}) = 1.67 \times 10^{5}\,\text{J}$.
Answer: $1.67 \times 10^{5}\,\text{J}$
Worked example: Ice to warm water
How much heat turns $0.20\,\text{kg}$ of ice at $-10\,^\circ\text{C}$ into water at $20\,^\circ\text{C}$?
- Warm the ice to $0\,^\circ\text{C}$: $(0.20)(2100)(10) = 4200\,\text{J}$.
- Melt it: $(0.20)(3.34 \times 10^{5}) = 66{,}800\,\text{J}$.
- Warm the water to $20\,^\circ\text{C}$: $(0.20)(4186)(20) = 16{,}744\,\text{J}$.
- Total: $4200 + 66{,}800 + 16{,}744 = 87{,}744\,\text{J} \approx 8.77 \times 10^{4}\,\text{J}$. Melting is three quarters of it.
Answer: $8.77 \times 10^{4}\,\text{J}$
Worked example: An ice cube in a drink
A $0.050\,\text{kg}$ ice cube at $0\,^\circ\text{C}$ goes into $0.30\,\text{kg}$ of water at $50\,^\circ\text{C}$ in an insulated cup. All the ice melts. What is the final temperature?
- Heat gained by the ice = heat lost by the water: $m_iL_f + m_ic\,T_f = m_wc(50 - T_f)$.
- $$(0.050)(3.34 \times 10^{5}) + (0.050)(4186)T_f = (0.30)(4186)(50 - T_f)$$ so $16{,}700 + 209.3\,T_f = 62{,}790 - 1255.8\,T_f$.
- $1465.1\,T_f = 46{,}090$, so $T_f = 31\,^\circ\text{C}$.
Answer: About $31\,^\circ\text{C}$
Common mistake
Forgetting the heat that melts the ice before the water can warm up. Leaving out $mL_f$ in the second example would give about $21{,}000\,\text{J}$, a quarter of the right answer. Check: list every stage between the start and the end, and include a $Q = mL$ term for each change of state you cross.