| This stew's
not cooking fast enough! We need to raise the temperature of the water
another 10° Celsius.
Let's see...20 kg of water and 70 kg of person, both with a specific heat capacity of 4.2 Joules/(gram-Celsius degree) would take...??? Help me Herb! How many Joules of heat energy will I need? By the way, the tub, being made out of special physics materials, does not have to be considered in the calculation. |
The quantity (Q) of energy involved
in a temperature change equals the product of specific heat capacity, mass,
and change in temperature. So,
Q = (4.2 J/gC°)(90x103g)(10C°)
|