Temperature Change of Apple Cobbler in Kitchen

Chrismb159
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hi, i am having trouble with this problem...

An apple cobbler is taken out of the oven at 7:00PM one Saturday night. At that time it is piping hot at 100C. At 7:10PM its temperature is 80C, and at 7:20pm, it is 65C. What is the temperature of the kitchen in which the apple cobbler is being held?
 
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What have you tried so far?
 
To solve you may use Newton's law of cooling

\int_{T_{0}}^{T} \frac{dT}{T-T_{s}} = - \kappa \int_{t_{0}}^{t} d t^{\prime}

where To is the initial temperature, Ts is the temperature of the environment (which is what we are after), k is a material constant, etc. Integrate both sides of this equation, and solve for T, and use the three boundary conditions to solve. You should get a reasonable answer for the room temperature. hope this helps, sincerely x
 
thank you for that, we recently learned about that in physics but for some reason i couldn't put 2 and 2 together, i was going in a completely different direction.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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