Dissolving some potassium bromide in 200cm3 of water leads to a decrease in temperature of 3oC. What will the heat released be?

1 Answer
Dec 19, 2014

The answer is q=2510J.

The amount of heat released can be calculated using the equation

q=mH2OcH2OΔT, where

mH2O - the mass of water;
cH2O -water's specific heat ( 4.18JgC);
ΔT - the change in temperature measured as TfinalTinitial.

A decrease in temperature will determine a negative ΔT, since this is equivalent to a lower final temperature (I'm assuming that the decrease in temperature was 3C, not 30C).

We can use water's density of approximately 1.00gcm3 to determine its mass

ρ=mVmH2O=ρV=1.00gcm3200.0cm3=200.0g

Therefore,

q=mcΔT=200.0g4.18JgC(3.00C)=2510J

The water lost energy in the form of heat equal to 2510J.