Question

Given 28.2 grams of an unknown substance, if the substance absorbs 2165 joules of energy and the temperature increases by 35 Kelvin, what is the specific heat of the substance?

2.14 x 106 J/g·K

2.19 x 100 J/g·K

5.73 x 10-4 J/g·K

2.69 x 103 J/g·K

Answer 1

They have given heat, change in temperature and mass, so
\[Q=mc \Delta T \\ \\ c=\frac{Q}{m \Delta T}\]

Answer 2

I honestly have no idea what to do...

Answer 3

It's a direct question just substitute in
Q=Heat
ΔT=Change in temperature
m=Mass
c=Specific heat capacity
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\[c=\frac{2165}{28.2(35)}=2.19J/gK\]

Answer 4

Thank youuu! :)