Question 98fe4

Mar 28, 2017

You will need to add 9.4 kJ of heat.

Explanation:

The energy $q$ required to heat an object is given by the formula

color(blue)(bar(ul(|color(white)(a/a)q = mcΔTcolor(white)(a/a)|)))" "

where

$m \textcolor{w h i t e}{l l}$ is the mass
$c \textcolor{w h i t e}{m l}$ is the specific heat capacity
ΔT is the change in temperature

$m \textcolor{w h i t e}{l l} = \text{27 g}$
$c \textcolor{w h i t e}{m l} = \text{4.184 J·°C"^"-1""g"^"-1}$
ΔT = "83 °C"

q = 27 color(red)(cancel(color(black)("g"))) × "4.184 J"·color(red)(cancel(color(black)("°C"^"-1"·"g"^"-1"))) × 83 color(red)(cancel(color(black)("°C"))) = "9400 J" = "9.4 kJ"#