Question

A body at a temperature of ${50}^{0}F$ is placed in an oven whose temperature is kept at ${150}^{0}F$. If after $10$ minutes the temperature of the body is ${75}^{0}F$, find the time required for the body to reach a temperature of ${100}^{0}F$ (in minutes).

• A. $24.09$
• B. $34.09$
• C. $44.09$
• D. $35.09$

Answer 1

Answer: (A)

$24.09$

Rate of heat gained is proportional to the difference between the surrounding's temperature and body's temperature,

i.e., $\frac{dT}{dt}=k(150-T)$

Where $T$ is the body's temperature and $k$ is the proportionality constant,

Separating the variables and integrating gives,

$\int \frac{1}{150-T}dT=\int kdt$

$\ln \left(\frac{1}{150-T}\right)=kt+c$ where $c$ is the integration constant,

Given after $10min$ the temperature of the body is ${75}^{0}F$

Substituting in the obtained equation gives,

$c=-\ln \left(75\right)-10k$

Substituting in the obtained equation gives,

$\ln \left(\frac{75}{150-T}\right)=k(10-t)$

Given at $t=0$ the temperature of the body is ${50}^{0}F$

Substituting this in the equation gives,

$k=\frac{\ln \left(0.75\right)}{10}$

Now to find the time required by the body to reach a temperature of ${100}^{0}F$

we need to substitute $T={100}^{0}F$ which gives,

$\ln 1.5=\frac{\ln \left(0.75\right)}{10}(10-t)$

This gives $t=24.0942min$