Question

A bowl of water has a temperature of 50◦C. It is put into a refrigerator where the temperature is 5◦C. After 0.5 hours, the water is stirred and its temperature is measured to be 20◦C. It is then left to cool further, take t in Newton’s law of cooling to be measured in minutes. Use Newton’s law of cooling to predict when the temperature will be 10◦C.

Answer 1

$T(t)=T_s+(T_o-T_s)e^{-kt}\\ T(t) \text{ --- temperature of the object at time t}\\ T_s \text{ --- temperature of surrounding}\\ T_o \text{ --- temperature of the object at time } t=0\\ k=\text{const}\\ \text{A bowl of water has a temperature of 50◦C. It is put}\\ \text{into a refrigerator where the temperature is 5◦C}.\\ \text{After 0.5 hours, the water is stirred and its temperature}\\ \text{is measured to be 20◦C. So we have:}\\ 20=5+(50-5)e^{-0.5k}\\ e^{-0.5k}=\frac{1}{3}\\ -(0.5)k=\ln\frac{1}{3}\\ k=\ln 9\\ \text{It is then left to cool further. We should predict when}\\ \text{the temperature will be 10◦C:}\\ 10=5+(20-5)e^{-(\ln 9) t}\\ e^{-(\ln 9) t}=\frac{1}{3}\\ t=0.5\\ \text{The temperature will be 10◦C in 30 minutes}\\ \text{(after we put a bowl of water into a refrigerator the second time)}.$