Question

A chemist mixes 75.0 g of an unknown substance at 96.5C w/ 1,150 g of water at 25.0C . if the final temperature of the system is 37.1C, what is the specific heat capacity of the substance ? Use 4.184 J /g Celsisus for the specific heat capacity of water.
a. 368
b. 13.1
c. 0.368
d. 0.0112

Answer 1

Remember: heat lost = heat gained 

When calculating heat loss or gain, remember 

mass*(spec heat cap)*(change in T) 

The unknown loses heat- we don't know the spec heat cap, so we'll call it x.

The water gains. I've omitted the units, but always use when solving problems on your own. 

75*x*(96.5-37.1) = 1150*4.184*(37.1-25) 

Now it's all set up- use algebra to get x, the spec heat cap of the unk in J/g*degC

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