3.6.2.1
Thermal equilibrium problems combine Q = mc delta theta and Q = mL
Thermal Energy Transfer — AQA A-Level Physics
- When hot and cold substances are mixed, energyThe capacity to do work. Measured in joules (J). flows from hot to cold until they reach the same final temperature (thermal equilibriumAn object is in equilibrium when the resultant force on it is zero. The object is either stationary or moving at constant velocityThe rate of change of displacement. A vector quantity. Measured in m s⁻¹..The state in which two objects in thermal contact have no net heat transfer between them because they are at the same temperature.).
- EnergyThe capacity to do work. Measured in joules (J). gained by the colder substance = energyThe capacity to do work. Measured in joules (J). lost by the warmer substance.
- If a phase change occurs (e.g. ice melting), the calculation has three stages: heat the solid to melting point, melt it, then heat the liquid to the final temperature.
$$Q_{\text{heat solid}} + Q_{\text{melt}} + Q_{\text{heat liquid}} = -Q_{\text{warm substance}}$$
Worked Example
200 g of coffee at 90 degrees C. Six ice cubes, each 14.5 g at -5 degrees C. $c_{\text{coffee}} = 4180$, $c_{\text{water}} = 4190$, $c_{\text{ice}} = 2090\,\text{J kg}^{-1}\,\text{K}^{-1}$. $L_{\text{fusion}} = 334\,000\,\text{J kg}^{-1}$. Find the final temperature.
Show Solution
1
Write the energy balance
Energy to heat ice to 0 degrees C + energy to melt ice + energy to heat meltwater to $T_{f}$ = energy lost by coffee cooling to $T_f$.
2
Set up equation
$$6m_i c_i (5) + 6m_i L + 6m_i c_w T_f = m_c c_c (90 - T_f)$$
3
Substitute and solve
- Left side constant terms: $6(0.0145)(2090)(5) + 6(0.0145)(334000) = 909 + 29058 = 29967$ J
- Left side T_f term: $6(0.0145)(4190)T_f = 364.5 T_f$
- Right side: $0.20(4180)(90 - T_f) = 75240 - 836 T_f$
- $29967 + 364.5 T_f = 75240 - 836 T_f$
- $1200.5 T_f = 45273$
- $T_f = 37.7$ degrees C
Answer
$T_f = 37.7$ degrees C
Examiner Tips and Tricks
- Always do a common-sense check.
- The final temperature must be between the two initial temperatures.
- If your answer falls outside this range, check your signs.