Answer:
[tex]T = 79 degree C[/tex]
Explanation:
As per Newton's Law of cooling we know that
[tex]\frac{dT}{dt} = k(T - T_s)[/tex]
now we have
temperature cool from 115 to 90 degree C in half an hour
so we have
[tex]T = \frac{115 + 90}{2} = 102.5 [/tex]
[tex]\frac{115 - 90}{0.5} = k (102.5 - 70)[/tex]
[tex]k = 1.54[/tex]
now we will say in next half an hour temperature changes to T
[tex]\frac{90 - T}{0.5} = 1.54(\frac{T + 90}{2} - 70)[/tex]
[tex]\frac{90 - T}{0.5} = 1.54(\frac{T - 50}{2})[/tex]
[tex]180 - 2T = 0.77T - 38.5[/tex]
[tex]T = 79 degree C[/tex]
The temperature at the end of another half hour is 78.85°.
From Newton's law of cooling;
T(t) = T(s) + (To - Ts)e^kt
Where;
T(t) = temperature at time t = 90°
T(s) = temperature of the surrounding = 70°
To = initial temperature = 115°
k = cooling constant
t = time taken = 30 mins
Substituting values;
90 = 70 + (115 - 70)e^30k
90 - 70 = (115 - 70)e^30k
20 = 45e^30k
20/45 = e^30k
0.444 = e^30k
ln0.444 = lne^30k
ln0.444 = 30k
k = ln0.444/30
k = -0.0271
At the end of another half hour, 60 minutes have now gone.
T(t) = 70 + (115 - 70)e^(-0.0271)(60)
T(t) = 78.85°
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