Question

The specific heat of a substance varies with temperatureaccording to c = 0.20 +0.99T + 0.040T²,with T in °C andc in cal/g·K. Findthe energy required to raise the temperature of 7.4 g of thissubstance from 7.0°C to 25°C.

Answer 1

For convenience, let us put,

                    c = 0.20 + 0.99T +0.040T² = a + bT+c.T²

For an infinitesimal rise in temperature, dT, theenergy required,

       dQ = m.c.dT = m( a +bT +c.T²).dT

          Q =Integral dQ

             =m.(a.T + 0.5bT² + 0.333c.T³).

             evaluated between T = 7 and T = 25

Hence, Q = 7.4 g(0.2 x18 + 0.997x288 + 0.0407x5094)cal/g

               = 7.4 x 498 = 3685cal

Answer 2

dQ = m * c * dT     where Q is in caloriesand T specified to be Celsius

                                                                                  25

So Q = m * [.20 * T + .99 * T^2 / 2 + .04 * T^3 /3]       = (523 - 30) * 7 = 3450cal

                                                                                  7

E = 4.19 J / cal * 3450 cal = 14,460 J