Discussion
Adding 50 mm of insulation of conductivity 0.04 W/mK to that wall raises its resistance to
Answer: C
Correct answer: 1.75 m2K/W
The insulation contributes \[ R = \frac{t}{k} = \frac{0.05}{0.04} = 1.25\ \mathrm{m^2K/W}, \] and resistances in series add: \( 0.5 + 1.25 = 1.75 \). The U-value falls from 2.0 to 0.57, so 50 mm of insulation has removed more than seventy per cent of the heat loss through that wall.
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