Specific Heat ... Please Help...?

Question

One way to test a metallic sample whether it is made of gold is to heat it up and place it in a bomb calorimeter to determine the specific heat of the sample. The specific heat of gold is .13J/g x K. What temperature change would prove that the metal was gold, if 15g of the sample at 99C were placed in a calorimeter initially at 25C. The heat capacity of the calorimeter is 25J/K.

Please Explain How to Solve...

Answer 1

The heat lost by the gold = the heat gained by the calorimeter

The heat lost by the gold = 0.13 J/g K x 15 g x (99 - T final)

The heat gained by the calorimeter = 25 J/K x (T final - 25)

0.13 J/g K x 15 g x (99 - T) = (T - 25) x 25 J/K

193.05 - 1.95T =25T - 625

818.05 = 26.95T

T = 30.35

Answer 2

let's find out the amount of excess energy the gold has because it is @ 99C instead of 25C , a difference of 74C or 74K:

dH = m C dT

dH =15 grams (0.13J/g-K) (74K)

dH = 144.3 Joules

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now let's assume that both the calorimeter & the gold are at 25 C ,... and we re-introduce that excess 144.3 joules:

dH = dH gold  & dH cal

dH = m C dT   &    C dT

144.3 = 15 g (0.13) dT   &    (25J/K) dT

144.3 = 1.95 dT   &    25 dT

144.3 =    26.95 dT

dT = 5.354 K  or C

your answer (2 sigfigs): 5.4 Celsius increase for the calorimeter(to 30.4C)

(or a decrease in temp of the metal of 68.6 Celsius to 30.4Celsius)