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1.00 kJ of heat is added to a slug of silver and a separate 1.00 kJ of heat is added to a slug of i

1.00 kJ of heat is added to a slug of silver and a separate 1.00 kJ of heat is added to a slug of iron. The heat capacity of the silver slug is 137J/°C while the heat capacity of the iron slug is 615 J/°C.


If the slugs are each originally at 25.00°C, what is the final temperature of each slug?


Explanation: To calculate heat, we can use the formula q=CΔT, where q is heat, C is heat capacity and ΔT = Tfinal - T initial.


Silver:

q=1.00kJ = 1000J (we had to convert kJ to J by multiplying by 1000) because heat capacity has units of joules

C silver = 137J/C

Tinital = 25.00°C

T final = ?


Let's plug everything into our formula:

q=CΔT

1000J = 137J/C (ΔT)

ΔT = 7.29927

ΔT = Tfinal - T initial.

7.29927 = Tfinal - 25.00°C

Tfinal = 32.299°C= 32.3°C



Iron:

q=1.00kJ = 1000J (we had to convert kJ to J by multiplying by 1000) because heat capacity has units of joules

C iron = 615 J/°C

Tinital = 25.00°C

T final = ?


Let's plug everything into our formula:

q=CΔT

1000J = 615 J/°C (ΔT)

ΔT = 1.626

ΔT = Tfinal - T initial.

1.626 = Tfinal - 25.00°C

Tfinal = 26.626°C= 26.6°C


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