Question

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

Answer 1

The answer is `q = -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`.