Question

A 2.46 g sample of a substance suspected of being pure gold is warmed to 72.7 ∘C and submerged into 15.0 g of water initially at 24.7 ∘C. The final temperature of the mixture is 26.6 ∘C.

Part A:

What is the specific heat capacity of the unknown substance?

Part B:

Could the substance be pure gold?

Answer 1

Heat lose of substance            =             heat gain of substance

mct                                       =              mct

2.46*C*(72.2-26.6)                  =               15*4.184*(26.6-24.7)

                    C                         = 1.063J/g-c

it is not pure gold

     pure gold heat capacity = 0.129J/g-c