Question

An 80.0-gram sample of water at 10.0°C absorbs 1680 Joules of heat energy. What is the final temperature of the water? a 50.0°C b 15.0°C c 5.00°C d 4.00°C

Answer 1

Answer:

b)15.0°C

Explanation:

Specific Heat of Water=4.2 J/g°C

This means, that 1 g of Water will take 4.2 J of energy to increase its temperature by 1°C.

∴80 g Water will take 80×4.2 J of energy to increase its temperature by 1°C.

80×4.2 J=336 J

Total Energy Provided=1680 J

The temperature increase=\frac{\textrm{Total energy required}}{\textrm{energy required to increase temperature by one degree}}

Temperature increase=$\frac{1680}{336}$

=5°C

Initial Temperature =10°C

Final Temperature=Initial + Increase in Temperature

=10+5=15°C