Question

question 10 (1 point)
a lab technician places some peanut butter into a bomb calorimeter. the heat capacity of the calorimeter is calibrated at 8.280 kj/°c. the temperature change of the calorimeter is 11 °c. what is the amount of thermal energy released by the sample of peanut butter?
91.0 kj
152 kj
381 kj
49 kj
223 kj
question 11 (1 point)

Explanation:

Step1: Recall the formula for heat energy

The formula for heat energy \( q = C\times\Delta T\), where \(C\) is the heat capacity of the calorimeter and \(\Delta T\) is the temperature change.

Step2: Substitute the given values

Given \(C = 8.280\space kJ/^{\circ}C\) and \(\Delta T=11^{\circ}C\). Then \(q=(8.280\space kJ/^{\circ}C)\times(11^{\circ}C)\)

Step3: Calculate the value

\(q = 8.280\times11=91.08\space kJ\approx91.0\space kJ\)

Answer:

91.0 kJ