SOLUTIONS TO EXERCISE 1


(a) The MKS unit for heat capacity uses Joules, kilograms and degrees Kelvin.
One calorie is equal to 4.2 Joule; one kg = 1000 grams; and one deg C equals
one degree K.  So

    1 cal/g/deg C  = (4.2 J/cal) * (1000 g/kg) * (1 deg C/deg K) * (1 cal/g/deg C)
                   = 4200 J/kg/deg K.

(b) dU = mCdT, where m = 1 kg, C = 4200 J/kg/deg K, and dT = (90 - 20) = 70 K.

    So dU = 1 * 4200 * 70 = 288,000 J.

(c) Since ethyl alcohol has a lower heat capacity than water, a mixture of ethyl
alcohol and water would have an average heat capacity lower than that of plain
water.  Thus, the same mass of vodka would require less heat energy to get the
same change in temperature.  If the heat output of the stove stays the same,
the vodka would get to 90 degrees C faster than an equal mass of plain water.
(By the way, the density of ethyl alcohol is about 0.79 kg/liter, so one kg of
vodka would actually have a slightly larger volume than one kg of water; make
sure your teapot is large enough.)