Temperature and Heat — Question 2

Question 2

A changing heat-transfer rate must be integrated

A body receives heat at rate $P(t)=120-20t$ watts for $0\le t\le8\,\mathrm s$, with $t$ measured in seconds. Negative $P$ means outward heat transfer. No other energy crosses the boundary.

1. Find the net heat transfer and the separate inward and outward amounts.

2. Find the time of maximum accumulated heat and the average signed rate over the full interval.

Question 2 - Solution

A changing heat-transfer rate must be integrated

(a) Method and calculation

$$Q(t)=\int_0^t(120-20t\prime)dt\prime=120t-10t^2.$$$$Q(8)=320\,\mathrm J,\quad Q_{\mathrm{in}}=Q(6)=360\,\mathrm J,\quad Q_{\mathrm{out}}=360-320=40\,\mathrm J.$$

(b) Interpretation and further calculation

$$\frac{dQ}{dt}=120-20t=0\ \Rightarrow\ t_{\max}=6.00\,\mathrm s.$$$$\overline P=\frac{Q(8)}8=40.0\,\mathrm W.$$The gross crossing amount is $360+40=400$ J; it differs from the signed net amount.

Physical check

After 6 s heat leaves, so accumulated heat decreases even though time continues to increase.