Question

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if a calorimeter has a heat capacity of $c_{cal}=1056 j^{circ}c^{-1}$, and a reaction caused a temperature increase of $3.7^{circ}c$, what are $q_{calorimeter}$ and $q_{chemicals}$?
$q_{calorimeter}=3.9 kj, q_{chemicals}=0 kj.$
$q_{calorimeter}=-3.9 kj, q_{chemicals}=-3.9 kj.$
$q_{calorimeter}=-3.9 kj, q_{chemicals}=3.9 kj.$
$q_{calorimeter}=3.9 kj, q_{chemicals}=3.9 kj.$
$q_{calorimeter}=3.9 kj, q_{chemicals}=-3.9 kj.$

Explanation:

Step1: Calculate heat absorbed by calorimeter

Use the formula $q_{cal}=C_{cal}\times\Delta T$. Given $C_{cal} = 1056\ J^{\circ}C^{-1}$ and $\Delta T=3.7^{\circ}C$.
$q_{cal}=1056\ J^{\circ}C^{-1}\times3.7^{\circ}C = 3907.2\ J=3.9\ kJ$

Step2: Determine heat of chemicals

According to the law of conservation of energy in a calorimetry - system, $q_{cal}+q_{chemicals}=0$. So $q_{chemicals}=-q_{cal}$. Since $q_{cal} = 3.9\ kJ$, then $q_{chemicals}=- 3.9\ kJ$.

Answer:

$q_{calorimeter}=3.9\ kJ,q_{chemicals}=-3.9\ kJ$ (the last option)