Hello, Bryan,
This asks what the temperature change would be if 10,500 Joules are added to 150 grams of water.
We need the specific heat of water (Cp) to determine this. Cp is a constant that can be written using several different units. For water:
| 4.186 J/gm*K |
1.00 cal/gm*K or Btu/lb*F |
75.2 J/mole*K |
There are others, but they are numerically equivalent when adjusted for the units.
In this case, we'll use 4.185 J/gm*K, since our inputs are both in Joules and grams. [One can use any of the others, but you would need to change the Joules into calories, grams into moles, etc. first.].
I'm lazy, so I'm sticking with J/gm*K.
4.186 is the amount of energy (Joules) it takes to raise 1 gram of water by 1 degree Kelvin. Just read the units as you say "I need 4.186 Joules to raise 1 gram of your friggin' water 1 degree Kelvin."
But you have 150 g, so just multiply:
4.186 J/g*K * 150 g = 627.9 Joules/K. The grams unit cancels. Read this as "I need 627,9 Joules to raise this loser's water sample by 1 degree Kelvin."
Now apply 10,500 Joules to the 150 grams (imagine 150g of water in a beaker over a hot plate that transfers 10,500 Joules only to the water).
I need to divide the total energy applied by the 627.9 Joules/K that we calculate for our sample.
= 10,500 J/(627.9 J/K) = 16.7 deg. K. [The Joules cancel and deg Kelvin moves to the top]
The temperature changes by 17 degrees Kelvin. I rounded off since we only have 2 sig figs (150g).
It goes up since we added energy. A negative 10,500 Joules would have reduced the temperature by 17 degrees, since the temperature value would then become a negative number.
The question does not specify the units of temperature, but you could change that to C or F. The increase in C would be the same (17) since the C and K scales have the same slope. C = K - 273.
I hope this helps,
Bob
