Question

how much energy is required to raise the temperature of a 300.0 gram block of lead from 22.3°c to 59.9°c? the specific heat of lead is 0.129 j/g°c.
round your answer to the nearest whole number.
a 1455 joules
b 1756 joules
c 1275 joules
d 1392 joules

Explanation:

Step1: Calculate the temperature change

$\Delta T=T_{final}-T_{initial}=59.9 - 22.3=37.6^{\circ}C$

Step2: Use the heat formula $Q = mc\Delta T$

Here, $m = 300.0$ g, $c=0.129$ J/g$^{\circ}C$, $\Delta T = 37.6^{\circ}C$
$Q=(300.0)\times(0.129)\times(37.6)$
$Q = 300\times0.129\times37.6$
First, $300\times0.129 = 38.7$
Then, $38.7\times37.6=(40 - 1.3)\times37.6=40\times37.6-1.3\times37.6$
$40\times37.6 = 1504$, $1.3\times37.6 = 48.88$
$1504-48.88=1455.12$

Answer:

A. 1455 joules