# How much heat is added if .685 g of water increases in temperature by 287 degrees C?

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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.