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

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

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Answer 1 · 2014-12-19T21:53:02

The answer is $q = - 2510 J$ $q = -2510J$ .

The amount of heat released can be calculated using the equation

$q = m_{H_{2} O} \cdot c_{H_{2} O} \cdot Δ T$ $q = m_(H_2O) * c_(H_2O) * DeltaT$ , where

$m_{H_{2} O}$ $m_(H_2O)$ - the mass of water;
$c_{H_{2} O}$ $c_(H_2O)$ -water's specific heat ( $4.18 \frac{J}{g^{\circ} C}$ $4.18 J/(g^@C)$ );
$Δ T$ $DeltaT$ - the change in temperature measured as $T_{f i n a l} - T_{i n i t i a l}$ $T_(fi nal) - T_(i nitial)$ .

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

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

$ρ = \frac{m}{V} \to m_{H_{2} O} = ρ \cdot V = 1.00 \frac{g}{c m^{3}} \cdot 200.0 c m^{3} = 200.0 g$ $rho = m/V -> m_(H_2O) = rho * V = 1.00 g/(cm^3) * 200.0 cm^3 = 200.0g$

Therefore,

$q = m \cdot c \cdot Δ T = 200.0 g \cdot 4.18 \frac{J}{g^{\circ} C} \cdot (- {3.00}^{\circ} C) = - 2510 J$ $q = m * c * DeltaT = 200.0g * 4.18 J/(g^@C) * (-3.00^@C) = -2510J$

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

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

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