A bar of metal is cooling from 1000°C to room temperature, 24°C. The temperature, H, of the bar t minutes after it starts cooling is given, in oc, by H = 24 + 976e-0.1t
(a) Find the temperature of the bar at the end of one hour. (Round your answer to the nearest degree.)
(b) Find the average value of the temperature over the first hour. (Round your answer to the nearest degree.)
(c) Is your answer to part (b) greater or smaller than the average of the temperatures at the beginning and the end of the hour? The answer to part (b) isecthan the average of the temperatures at the beginning and the end of the hour, because the graph is concave

The temperature of the bar at the end of one hour is [tex]26.4192^\circ[/tex] and the average value of the temperature over the first hour is [tex]186.7622^\circ[/tex].

Given :

A bar of metal is cooling from 1000°C to room temperature, 24°C.
The temperature, H, of the bar t minutes after it starts cooling is given, in [tex]\rm ^\circ C[/tex], by H = 24 + 976[tex]\rm e^{-0.1t}[/tex].

a) The temperature of the bar at the end of one hour is given by:

H = 24 + 976[tex]\rm e^{-0.1t}[/tex]

Now, put the value of t = 60 min in the above equation.

[tex]\rm H = 24 +( 976\times e^{-0.1\times 60})[/tex]

[tex]\rm H = 24 + 976\times e^{-6}[/tex]

H = 24 + 2.4192

[tex]\rm H = 26.4192^\circ[/tex]

b) To find the average value of the temperature over the first hour, integrate the given expression.

[tex]\rm \int\limits^{60}_0 {H} \, dt =24 +\int\limits^{60}_0 {\dfrac{976}{60}e^{-0.1t}} \, dt[/tex]

By solving the above integration, the value of H is:

H = [tex]186.7622^\circ[/tex]

c) The answer in part b) is smaller than the average of the temperature at the beginning.

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