How many grams of silver are needed to absorb 1.00 kJ of energy changing the temperature 33.0 C to 41.8°C?

Apr 14, 2017

473.5 grams

Explanation:

$q = m c \Delta T$

q = heat energy (joules)
m = mass (grams)
c = specific heat ($\text{Joules"/"gram * Celsius}$)
$\Delta T$ = change in temperature

In this case:
q = 1 KJ = 1000 J
m = ?
c = 0.24
$\Delta T$ = 41.8 - 33 = 8.8

$q = m c \Delta T$
1000 = (m)(0.24)(8.8)
1000 = (m)(2.112)
m = 473.5