Question

31.5 grams of an unknown substance is heated to 102.4 C and then placed into a calorimeter containing 103.5 grams of water at 24.5 C. If the final temperature reached in the calorimeter is 32.5 C, what is the specific heat of the unknown substance?

Answer 1

`0.38 C g^-1 (degree C)^-1`

Explanation:

Since heat gained by substance equals heat lost by water, and heat transferred is given by
`H = m*c*Delta (theta)`
where, m is the mass of the substance,
c is its specific heat capacity
And `Delta (theta)` is change in temperature,
`31.5*c*(102.4-32.5)=103.5*1*(32.5-24.5)`
Then,
`c=(103.5*8)/(31.5*69.9)`
`=828/2201.85`
`=0.38`