Question

A body with an initial temperature ${\theta }_1$ is allowed to cool in surrounding which is at a constant temperature of ${\theta }_0({\theta }_0<{\theta }_1)$, Assume that Newton's law of cooling is obeyed. Let $k=constant$. The temperature of the body after time $t$ is best expressed by

• A. $({\theta }_1-{\theta }_0)e^{kt}$
• B. $({\theta }_1-{\theta }_0)ln\left(kt\right)$
• C. $({\theta }_0+({\theta }_1-{\theta }_0\left)e^{-kt}\right)$
• D. ${\theta }_1{e}^{-kt}-{\theta }_0$

Answer 1

Answer: (C)

$({\theta }_0+({\theta }_1-{\theta }_0\left)e^{-kt}\right)$

By newton's law of cooling, the rate of change in temperature is directly proportional to difference in temperature of body and surrounding. i.e., $d\theta /dt=-k(\theta -{\theta }_o)$ $\Rightarrow d\theta /(\theta -{\theta }_o)=-kdt$ $\Rightarrow {\int }_{{\theta }_1}^{\theta }\frac{1}{(\theta -{\theta }_o)}=-k{\int }_0^{t}dt$ $\Rightarrow log(\theta -{\theta }_o/{\theta }_1-{\theta }_o)=-kt$ $\Rightarrow \theta ={\theta }_o+({\theta }_1-{\theta }_o)e^{-kt}$