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)