Question

A 65.0 gram sample of some unknown metal at 100.0° C is added to 177.1 grams of water at 24.5° C. The temperature of the water rises to 27.0° C. If the specific heat capacity of liquid water is 4.18 J/ (°C × g), what is the specific heat of the metal?

Answer 1

0.285 J/ (°C × g).

0.390 J/ (°C × g).

1.05 J/ (°C × g).

2.56 J/ (°C × g).
These are the answer choices. Its for my Ap chem packet due when I get back to school. Can't figure this one out

Answer 2

if:
Q = m cp (T2-T1) or
\[dQ = m \int\limits c _{p} dt\]
so with known water capacity you calculate the heath needed for temperature to rise to that defined final temperature and after that you use same heat as an value to calculate heat capacity of metal in other words:
w - index of water
m - index of metal
Qw = mw cpw DTw
so
Qw=Qm
Qw = mm cpm DTm
cpm = Qw/ mm DTm
where DT is delta or difference in temperature ex 100°C-50°C = 50 °C
that is so cause same heat "apsorbed" by water is same heat released by metal...
hope you get it, if not just write part you are unclear about ;)