Specific Heat?

When 100.0g of a metal at 100.0oC is placed in 100.0g of water initially at 20.0oC the final temperature of the metal and the water is 32.0oC. Given that the specific heat of water is 1.00 cal/gram-degree calculate the specific heat of the metal in cal/g-degree.

Answer:
Start with the equation of sp heat. Sp heat=cal/(gm*kelvins)

So the amount of heat energy is the constant that doesnt change throughout the whole experiment

that energy=sp heat(gm)(kelvins)

Now convert our temps to kelvins... 20c is 293k, 100c is 373k

So for water, the energy=1(100)(293)=29300cal
For the metal, energy=spHeat(100)(373)=spHeat...

So total energy=spHeat(37300)+29300

Now at the end of the experiment (32c=305K):

water's energy=30500
metal's energy=spHeat(30500)

total energy=spHeat(30500)+30500

Now we have two equations we can solve... algebra now. Let's use S for the specific heat of the metal and E for total energy of the system

E=s(30500)+30500
E=s(37300)+29300

s(37300)+29300=s(30500)+30500
s(6800)=1200

s=~0.76

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