Question

when 19.55 g of solid k dissociates in 500.0 g of water in a coffee cup calorimeter, the temperature increases from 20.0 °c to 61.5 °c.
$c_{sol} = 4.18 \\, \text{j/g °c}$
$\text{mass}_{sol} = 519.6 \\, \text{g}$
what is the the heat of solution, $q_{sol}$?
$q_{sol} = ? \\, \text{j}$
enter either a + or - sign and the magnitude.

Explanation:

Step1: Recall the formula for heat

The formula to calculate heat is \( q = mc\Delta T \), where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the change in temperature.

Step2: Calculate the change in temperature

\( \Delta T = T_{final} - T_{initial} = 61.5^{\circ}C - 20.0^{\circ}C = 41.5^{\circ}C \)

Step3: Plug in the values into the formula

We know \( m = 519.6 \, g \), \( c = 4.18 \, J/g^{\circ}C \), and \( \Delta T = 41.5^{\circ}C \). So, \( q_{sol} = 519.6 \, g \times 4.18 \, J/g^{\circ}C \times 41.5^{\circ}C \)
First, calculate \( 519.6 \times 4.18 = 519.6\times4 + 519.6\times0.18 = 2078.4 + 93.528 = 2171.928 \)
Then, multiply by 41.5: \( 2171.928\times41.5 = 2171.928\times(40 + 1.5) = 2171.928\times40 + 2171.928\times1.5 = 86877.12 + 3257.892 = 90135.012 \)
Since the temperature of the solution increased, the heat of the solution is positive (the solution absorbed heat, or the reaction released heat which was absorbed by the solution, so \( q_{sol} \) is positive as the solution's energy increased).

Answer:

\( +90100 \) (rounded to a reasonable number of significant figures, or \( +90135 \) if we keep more precision)