Question

what quantity of heat (in kj) will be absorbed by a 39.5 g piece of aluminum (specific heat = 0.930 j/g·°c) as it changes temperature from 23.0 °c to 67.0 °c? answer: kj

Explanation:

Step1: Calculate temperature change

$\Delta T=T_{final}-T_{initial}=67.0 - 23.0=44.0^{\circ}C$

Step2: Use heat formula

The formula for heat $q = mc\Delta T$, where $m = 39.5g$, $c=0.930J/g\cdot^{\circ}C$, $\Delta T = 44.0^{\circ}C$
$q=(39.5g)\times(0.930J/g\cdot^{\circ}C)\times(44.0^{\circ}C)$
$q = 39.5\times0.930\times44.0J$
$q=39.5\times40.92J$
$q = 1616.34J$

Step3: Convert to kJ

Since $1kJ = 1000J$, then $q=\frac{1616.34J}{1000}=1.62kJ$ (rounded to three significant figures)

Answer:

$1.62$