3.6.2.1

Thermal equilibrium problems combine Q = mc delta theta and Q = mL

Thermal Energy Transfer — AQA A-Level Physics

$$Q_{\text{heat solid}} + Q_{\text{melt}} + Q_{\text{heat liquid}} = -Q_{\text{warm substance}}$$
Worked Example
200 g of coffee at 90 degrees C. Six ice cubes, each 14.5 g at -5 degrees C. c_coffee = 4180, c_water = 4190, c_ice = 2090 J kg^{-1} \(K^{-1}\). L_fusion = 334 000 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.
Thermal Energy Transfer Overview