A 0200 kg sample is placed in a cooling apparatus that remov

A 0.200 kg sample is placed in a cooling apparatus that removes energy as heat at a constant rate of 3.470 W. The figure gives the temperature T of the sample versus time t. The temperature scale is set by T_s = 70 degree C and the time scale is set by t_s = 30 min. What is the specific heat of the sample? Give answer in J/(kg middot C degree).

Solution

Energy removed to lower the temp by deltaT, Q = m C deltaT

and rate of energy = Q/t

Q/t = m C deltaT/t

deltaT / t = 3.470 / m C

deltaT = (3.470 / (0.20 x C))t

hence slope from equation is = 17.35 /C

from figure, slope = 2.5 Ts / 2ts = 1.25 ( 70 / (30 x 60 sec))

= 0.0486

so, 17.35 / C = 0.0486

C = 356.9 J/kg C