Question

A bar of metal is cooling from 1000°C to room temperature, 22°C. The temperature, H, of the bar t minutes after it starts cooling is given, in °C, by H = 22 + 978e−0.1t.

Answer 1

Answer:
       I assume you're looking for a time-weighed average rather than just the [H(0) + H(60)]/2 average.
       Havg = (1/60)â«[20+980e^(-0.1t)]dt (evaluated from 0 to 60)
       Havg = (1/60)[20t - 98e^(-0.1t)] evaluated from 0 to 60
       Havg = (1/60)[20(60) - 98e^(-6) - 20(0) + 98e^(0)] = 1297.76/60 = 21.629 °C