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 = -2510Jq=2510J.

The amount of heat released can be calculated using the equation

q = m_(H_2O) * c_(H_2O) * DeltaT, where

m_(H_2O) - the mass of water;
c_(H_2O) -water's specific heat ( 4.18 J/(g^@C));
DeltaT - the change in temperature measured as T_(fi nal) - T_(i nitial).

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

We can use water's density of approximately 1.00g/(cm^3) to determine its mass

rho = m/V -> m_(H_2O) = rho * V = 1.00 g/(cm^3) * 200.0 cm^3 = 200.0g

Therefore,

q = m * c * DeltaT = 200.0g * 4.18 J/(g^@C) * (-3.00^@C) = -2510J

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