Question #98fe4

1 Answer
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

#mcolor(white)(ll)# is the mass
#c color(white)(ml)# is the specific heat capacity
#ΔT# is the change in temperature

In your problem,

#m color(white)(ll)= "27 g"#
#c color(white)(ml)= "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"#