Question

question 13
calculate the amount of heat, in kj, associated with 127 g of mercury cooling from 124.8°c to 74.8°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: Determine the formula

The formula for heat \(q = mc\Delta T\), where \(m\) is mass, \(c\) is specific heat, \(\Delta T=T_{final}-T_{initial}\)

Step2: Substitute the values

Given \(m = 127g\), \(c=0.139J/g^{\circ}C\), \(T_{initial}=124.8^{\circ}C\), \(T_{final}=74.8^{\circ}C\). Then \(\Delta T=74.8 - 124.8=- 50^{\circ}C\)
\(q=(127g)\times(0.139J/g^{\circ}C)\times(-50^{\circ}C)\)
\(q=127\times0.139\times(- 50)J\)
\(q=-880.15J\)
Convert to \(kJ\): \(q=-0.88015kJ\)

Answer:

\(-0.88kJ\)