Question

calculate the amount of heat, in kj, associated with 169 g of mercury cooling from 104.4 °c to 58.3 °c. note that the specific heat of mercury is 0.139 j/g °c. report your answer to two places past the decimal point.

Explanation:

Step1: Calculate the temperature change

The formula for temperature change $\Delta T=T_{final}-T_{initial}$. Here, $T_{initial} = 104.4^{\circ}C$ and $T_{final}=58.3^{\circ}C$. So, $\Delta T=58.3 - 104.4=- 46.1^{\circ}C$

Step2: Use the heat - transfer formula

The heat - transfer formula is $q = mc\Delta T$, where $m$ is the mass, $c$ is the specific heat capacity, and $\Delta T$ is the temperature change. Given $m = 169g$, $c=0.139J/g^{\circ}C$, and $\Delta T=-46.1^{\circ}C$.

Substitute the values into the formula: $q=(169g)\times(0.139J/g^{\circ}C)\times(- 46.1^{\circ}C)$

First, calculate $(169\times0.139)=23.491$. Then, $23.491\times(-46.1)=-1082.9351J$

Step3: Convert joules to kilojoules

Since $1kJ = 1000J$, then $q=\frac{- 1082.9351J}{1000}=-1.0829351kJ$

Answer:

$-1.08kJ$