## Essential University Physics: Volume 1 (4th Edition)

We divide the value of Q by the rate at which heat is absorbed to find: $\Delta t = \frac{ \Delta Q}{10^5} = \frac{75 \times 2000 \times .2 \times 30}{10^5 } = 9 \ hours$ For the wood house: $\Delta t = \frac{ \Delta Q}{10^5} = \frac{15 \times 2000 \times .33 \times 30}{10^5 } = 3 \ hours$