Question

how much heat in kilojoules must be absorbed by 375 grams of water to raise its temperature from 15.0 °c to 40.0 °c?
35.0 kj
39.2 kj
-39.2 kj
375 kj

Explanation:

Step1: Calculate the temperature change

$\Delta T=T_{final}-T_{initial}=40.0^{\circ}C - 15.0^{\circ}C=25.0^{\circ}C$

Step2: Use the specific - heat formula

The specific heat capacity of water $c = 4.18\space J/(g\cdot^{\circ}C)$. The formula for heat is $q = mc\Delta T$, where $m = 375\space g$, $c=4.18\space J/(g\cdot^{\circ}C)$ and $\Delta T = 25.0^{\circ}C$.
Substitute the values: $q=(375\space g)\times(4.18\space J/(g\cdot^{\circ}C))\times(25.0^{\circ}C)$
$q = 375\times4.18\times25\space J$
$q=375\times104.5\space J$
$q = 39187.5\space J$

Step3: Convert joules to kilojoules

Since $1\space kJ=1000\space J$, then $q=\frac{39187.5\space J}{1000}=39.1875\space kJ\approx39.2\space kJ$

Answer:

39.2 kJ