Question

The specific heat of liquid ethyl alcohol is 2.42 j/g C and its density of 0.7893 g/mL. A piece of solid sliver (specific heat 0.24 J/g C) was heated to
95 C and placed in 25.6 mL of ethyl alcohol that had a temp of 19.27 C. The final temp of alcohol and silver was 23.5 C. How many grams of silver were used?

Answer 1

First off, as any good chemist would tell you, we have to know if this is an isolated system or not, or else there could be energy (namely heat) that could escape the system or matter from evaporation and stuff.

SOLUTION

25.6 (volume) * 0.7893 (density) = mass of the ethyl alcohol 
If it was heated to 19.27 degrees Celsius, this translates to 292.27 Kelvin. The specific heat of liquid ethyl alcohol is 2.42 j/(g*C), which is the same as 2.42 j/(g*K), therefore it has in total 2.42 * 292.27 * 25.6 * 0.7893 Joules of heat.

The silver has x grams, and its specific heat is 0.24 J/g*C = 0.24 J/g*K. If it was heated up to 95 degrees, that means it was heated to 368 degrees Kelvin. Therefore it 368 * 0.24 * x Joules of heat.

If the final temperature was 23.5 degrees Celsius of the entire thing, than the ethyl alcohol had 2.42 * 296.5 * 25.6 * 0.7893 Joules of heat. Getting the difference of the first state and the second state, we get 2.42 * (296.5 - 292.27) * 25.6 * 0.7893 Joules. This was added to the silver, which now has 296.5 * 0.24 * x Joules of heat. Subtracting the first state from the second state we get (296.5-368) * 0.25 * x Joules. The absolute value of this equals 2.42 * (296.5 - 292.27) * 25.6 * 0.7893 Joules. Doing all the math out, 206.84155853 = 17.16x.

Therefore x = 12.05 grams.

There are 12.05 grams of silver.

Answer 2

MAg*cAg*(T1-T)=ρalc*Valc*calc*(T-T2)
mAg=?(g)
cAg=0.24J/gC
T1=95
T=23.5
Valc=25.6ml
ρalc=0.7893g/ml
T2=19.27
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