We submerge a 32.6-g copper weight, initially at 71.5 ∘C into 64.1 g of water at...

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We submerge a 32.6-g copper weight, initially at 71.5 ∘C into 64.1 g of water at...

We submerge a 32.6-g copper weight, initially at 71.5 ∘C into 64.1 g of water at 23.0 ∘C in an insulated container. What is the final temperature of both substances at thermal equilibrium?

A 3.00-g sample of a substance is warmed to 83.7 ∘C and submerged into 15.2 g of water initially at 23.7 ∘C. The final temperature of the mixture is 25.4 ∘C. What is the heat capacity of the unknown substance?

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Answer 1

Answer #1

a. specific heat capacity of Cu = 0.385J/g-⁰C

heat capacity of H2O = 4.184J/g-⁰C

heat lose of copper = heat gain of water

mc $\Delta$ T = mc $\Delta$ T

32.6*0.385*(71.5-t) = 64.1*4.184*(t-23)

t = 25.17⁰C

the final temperature of both substances at thermal equilibrium = 25.17⁰C

b.

heat lose of substance = heat gain of water

mc $\Delta$ T = mc $\Delta$ T

3*c*(83.7-25.4) = 15.2*4.184*(25.4-23.7)

c =0.62J/g-⁰C

The heat capacity of the unknown substance c =0.62J/g-⁰C

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Answer 2

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Add Answer to: We submerge a 32.6-g copper weight, initially at 71.5 ∘C into 64.1 g of water at...

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