A chemist mixes 75.0 g of an unknown substance at 96.5°C with 1150 g of water at 25.0°C.?
If the final temperature of the system is 37.1°C, what is the specific heat capacity of the substance? Use 4.184 J / g °C for the specific heat capacity of water
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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