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

Dec 19, 2014

The answer is $q = - 2510 J$.

The amount of heat released can be calculated using the equation

$q = {m}_{{H}_{2} O} \cdot {c}_{{H}_{2} O} \cdot \Delta T$, where

${m}_{{H}_{2} O}$ - the mass of water;
${c}_{{H}_{2} O}$ -water's specific heat ( $4.18 \frac{J}{{g}^{\circ} C}$);
$\Delta T$ - the change in temperature measured as ${T}_{f i n a l} - {T}_{i n i t i a l}$.

A decrease in temperature will determine a negative $\Delta T$, since this is equivalent to a lower final temperature (I'm assuming that the decrease in temperature was ${3}^{\circ} C$, not ${30}^{\circ} C$).

We can use water's density of approximately $1.00 \frac{g}{c {m}^{3}}$ to determine its mass

$\rho = \frac{m}{V} \to {m}_{{H}_{2} O} = \rho \cdot V = 1.00 \frac{g}{c {m}^{3}} \cdot 200.0 c {m}^{3} = 200.0 g$

Therefore,

$q = m \cdot c \cdot \Delta T = 200.0 g \cdot 4.18 \frac{J}{{g}^{\circ} C} \cdot \left(- {3.00}^{\circ} C\right) = - 2510 J$

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