Question

A body cools from 90°C to 80°C in 5 minutes. Under the same external conditions to cool from 80°C to 70°C the body will take
1. 5 minutes
2. 4 minutes
3. 2.5 minutes
4. More than 5 minutes

Answer 1

Correct Answer - Option 4 : More than 5 minutes

Concept:

Heat transfer rate in transient heat transfer

\(\frac{{{T_{io}} - {T_\infty }}}{{T - {T_\infty }}} = {e^{\frac{{hA}}{{\rho V{C_p}}}t}}\)

Here h, A, ρ, V, and C_p are constant.

So, \(\frac{{{\rm{hA}}}}{{\rho V{c_p}}} = B\;\)

Now \(\frac{{{T_{io}} - {T_\infty }}}{{T - {T_\infty }}} = {e^{Bt}}\)

Here T_io is the initial temperature at t = 0 and T is temperature of body at any instant.

\(\begin{array}{l} \frac{{90 - {T_\infty }}}{{80 - {T_\infty }}} = {e^{B \times 300}}\\ \frac{{80 - {T_\infty }}}{{70 - {T_\infty }}} = {e^{Bt}} \end{array}\)

Let T_∞ = 0° C

\(\begin{array}{l} \frac{9}{8} = {e^{300\;B}}\\ \frac{{\ln \left( {\frac{9}{8}} \right)}}{{300}} = B\\ \frac{8}{7} = {e^{Bt}} \end{array}\)

B = 3.92 × 10^-4

\(\begin{array}{l} \ln \left( {\frac{8}{7}} \right) = Bt\\ \Rightarrow \ln \left( {\frac{8}{7}} \right) = \frac{{\ln \left( {\frac{9}{8}} \right) \times t}}{{300}}\\ t = \frac{{\ln \left( {\frac{8}{7}} \right)}}{{\ln \left( {\frac{9}{8}} \right)}} \times 300 = 340\;sec > 5~minutes \end{array}\)