Question

A 20.00 g sample of metal is warmed to 165˚C in an oil bath. The sample is then transferred to a coffee cup calorimeter that contains 125.0 g of water at 5.0˚C. The final temperature of the water is 8.8˚C.
mrtal water
m=20g m=125g
t1=165c t1=5
c=? t2=8.8
c=4.18j/g.c
qtot=qmetal + qwater
=mct + mct
= (20g)(c)(x-165)+(125g)(8.8-5)(4.18)


please help

Answer 1

It isn't clear what you want? I assume that is you want the specific heat of the metal. Here is your work.
= (20g)(c)(x-165)+(125(8.8-5(4.18)

<b>Just a change or two.
1. Make that = 0.
2. x is 8.8. If the final T is 8.8, that is not only the final T for the water but also the final T for the metal.
3. Solve for c.</b>