Question

A solid copper ball of mass 500 gm when quenched in a water bath at 30°C, cools from 530°C to 430°C in 10 sec. What will be the temperature of the ball after the next 10 seconds?

1. 300°C
2. 320°C
3. 350°C
4. Cannot be determine

Answer 1

Correct Answer - Option 3 : 350°C

Concept

This is a case of an unsteady-state heat transfer problem.

In order to obtain the temperature after t time the formula is given by

\(\frac{{T - {T_∞ }}}{{{T_i} - {T_{∞ \;}}}} = {\rm{exp}}\left( { - \frac{{hA}}{{ρ VC}} \times τ } \right)\)

where T = temperature after time t, T_∞ = temperature of bath or environment, T_i = initial temperature, τ is the time,

h is heat transfer coefficient, A is the area of cross-section, ρ is the density of ball, V is the volume of ball, C is the heat capacity.

Calculation:

Given: 

T_∞ = 30°C, Ti = 530°C , T = 430°C ,τ = 10 sec .

\(\frac{{T - {T_∞ }}}{{{T_i} - {T_{∞ \;}}}} = {\rm{exp}}\left( { - \frac{{hA}}{{ρVC}} \times τ } \right)\)

putting the value,

\(\frac{{430 - 30}}{{530 - 30}} = {\rm{exp}}\left( { - \frac{{hA}}{{ρVC}} \times 10} \right)\)

\(\frac{{hA}}{{ρVC}}\) = 0.022314

Now temperature of the ball after the next 10 sec means the total time is 20 sec (τ = 20 sec)

Note: Time is always taken from initial.

\(\frac{{T - {T_∞ }}}{{{T_i} - {T_{∞ \;}}}} = {\rm{exp}}\left( { - \frac{{hA}}{{ρVC}} \times τ } \right)\)

\(\frac{{T - 30}}{{530 - 30}} = {\rm{exp}}\left( { - \frac{{hA}}{{ρVC}} \times 20} \right)\)

put the value of \(\frac{{hA}}{{ρVC}}\) from the above 

\(\frac{{T - 30}}{{530 - 30}} = {\rm{exp}}\left( { - 0.022314 \times 20} \right)\)

T - 30 = 320

T = 350°C.