How much heat is added if .685 g of water increases in temperature by 287 degrees C?
The temperature of the water will increase in proportion to the amount of heat added. The quantity of heat that is needed can be calculated by the following formula:
heat added (or lost) = mass of water x specific heat of water x change in temperature
where specific heat is the amount of heat required to raise the temperature of a unit mass of a substance by 1 degree Celsius. In case of water, the specific heat is 4.186 J/K/g.
However, for a temperature change of 287 degrees, one must note that water converts to steam at 100 degrees C and the specific heat of steam is 1.996 J/K/g and the latent heat of vaporization is 2256 J/g.
Assuming some initial temperature (since it is not given), say 20 degrees C. Thus, the water sample undergoes three phases:
Boiling from 20 degrees C to 100 degrees C (80 degree Temperature change):
Heat needed = 0.685 g x 4.186 J/K/g x (100 -20) = 229.4 J
Phase change to steam at 100 degrees C:
Heat needed = mass x latent heat = 0.685 g x 2257 J/g = 1546.1 J
Heating steam from 100 degrees C to 307 degrees (temperature change by 207 degrees C):
Heat needed = 0.685 x 1.996 x (307 - 100)
= 283 J
Thus, the total heat needed to heat 0.685 gm of water, from 20 degrees C (assumed initial temperature) to 307 degrees C (an increase of 287 degrees C) is:
229.4 + 1546.1 + 283 J = 2058.5 J
Hope this helps.