Question

calculate the amount of heat, in kj, associated with 212 g of mercury cooling from 160.9 °c to 45.4 °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, \(T_{initial} = 160.9^{\circ}C\) and \(T_{final}=45.4^{\circ}C\), so \(\Delta T=45.4 - 160.9=- 115.5^{\circ}C\)

Step2: Use the heat formula \(q = mc\Delta T\)

Given \(m = 212g\), \(c=0.139J/g^{\circ}C\), and \(\Delta T=-115.5^{\circ}C\). Substitute the values into the formula: \(q=(212g)\times(0.139J/g^{\circ}C)\times(- 115.5^{\circ}C)\)

First, calculate \((212\times0.139)=29.468\), then \(29.468\times(-115.5)=-3403.554J\)

Step3: Convert joules to kilojoules

Since \(1kJ = 1000J\), then \(q=-3403.554J=- \frac{3403.554}{1000}kJ=-3.403554kJ\)

Answer:

\(-3.40kJ\)