The temperature of an object increases by 25.9 °C when it absorbs 3703 J of heat. Calculate the heat capacity of the object. С — J/°C The mass of the object is 371 g. Use the table of specific heat values to identify the composition of the object. Substance Specific heat (J/(g °C)) gold 0.129 сopper 0.385 iron 0.444 aluminum 0.900 The object is composed of gold.

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The temperature of an object increases by 25.9 °C when it absorbs 3703 J of heat. Calculate the heat capacity of the object.
C =
J/°C
The mass of the object is 371 g. Use the table of specific heat values to identify the composition of the object.
Substance
Specific heat (J/(g °C))
gold
0.129
сopper
0.385
iron
0.444
aluminum
0.900
The object is composed of
O gold.
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Transcribed Image Text:of 10 The temperature of an object increases by 25.9 °C when it absorbs 3703 J of heat. Calculate the heat capacity of the object. C = J/°C The mass of the object is 371 g. Use the table of specific heat values to identify the composition of the object. Substance Specific heat (J/(g °C)) gold 0.129 сopper 0.385 iron 0.444 aluminum 0.900 The object is composed of O gold. contact us help careers privacy policy terms of use about us
Expert Solution
Step 1

Given data, heat absorbed=3703 J.

Temperature of object increases by 25.9℃.

We know ,Heat capacity of an object=  Heat abosorbed / change in temperature

C= Q/∆T=3703J ÷ 25.9 ℃=142.97 J/℃.

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