A 5.82-kg piece of copper metal is heated from 21.5°C to 328.3°C. What is the heat absorbed (in kJ) by the metal?

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Answer 1 · 2017-07-18T07:37:45

The heat absorbed is #=687.4kJ#

The heat absorbed is

#E=m*C*DeltaT#

The mass is #=5.82kg#

The specific heat of copper is #C=0.385kJkg^-1ºC^-1#

The change in temperature is

#DeltaT=328.3-21.5=306.8#

Therefore,

#E=5.82*0.385*306.8=687.4kJ#

Answer 2 · 2017-07-18T08:01:23

Here are all informations that we have :

Mass of copper(m) #=5.82g#
Specific Heat Capacity (I will just name it SHC) of copper #=0.385 J(g°C)^-1# #rarr# simpler way to write 0.385 J/g°C
Change in temperature (#DeltaC#) #=328.3°C-21.5°C=306.8°C#

Then we start

Firstly the formula,

SHC #=(Heat)/(m*DeltaC)#

Then we replace the values and calculate it

#0.385=(Heat)/(5.82*306.8°C)#

#Heat=0.385*(5.82*306.8°C)#

Therefore, heat absorbed#=687.44676 J=687(3s.f)#

(I put the answer at 3 significant figures (s.f) but for further calculations use the first value obtained)