To find the heat capacity is q=mcΔT..How do i find the change in temperature?

Question

The question is like this: The heat required in an operation was 1000J.The operation involves heating of 1g water from 30 deg to a higher temperature.Calculate the higher temperature?

Answer 1

tlbs is correct. The final temp would be 100 C. Any phase change is isothermal.

Answer 2

What Doctor Q says is basically correct. The only strange part of this problem, is that 1000J applied to 1 g of water (at 30 degrees C) is alot of energy and might change it to steam.

The energy applied to the water is "linear" to go from 30 degrees to 100 degrees. When you reach the boiling point, there is another amount of energy called the latent heat of vaporization that is necessary to form steam. This makes the formula q=m*c*deltaT, non-linear.

The latent heat of vaporization for water is 2260 kJ/kg or 2260 J/g

Since there were only 1000 J of energy to begin with and you need over twice as much to get the water to boil, the final temperature of the water would be 100 degrees C, with the water just starting to boil.

Here is a real good explanation (with graphics):
http://www.physchem.co.za/Heat/Latent.htm

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Answer 3

Specific heat of water = 4.186 joule/gram °C = 4186 Joules per Kg

Re-arrange you equation for ΔT -:

ΔT = q / mc

ΔT = 1000 / 1 x 10^-3 x 4186
ΔT = 238.892 Degrees
Remember this is the 'change in temperature' so we have to add on the original 30 degrees, which gives the new final higher temp of -: 268.892 degrees celsius.

Hope this helps....