If 70 Btu of heat energy are added to 2.5 lb of water at 35F, what will be the final temperature of the water? Use formula Q = mC(Tf – Ti), and solve for Tf
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3 Answers

Q = heat
m = mass
C = specific heat
Tf = final temperature
Ti = initial temperature
So...
70 Btu = 2.5 lbs(covert to mass) * specific heat of water * (Tf  35)
If we assume the 2.5 lbs of water is lbm (mass), then:
70 = 2.5(1)(Tf  35)
70 = 2.5(Tf)  2.5(35)
70 = 2.5Tf  87.5
2.5Tf = 157.5
Tf = 157.5/2.5
Tf = 63 degF
Best of luck!

Q = mC(TfTi)
Q = 70 Btu
m = 2.5 lb
C of water = 1 Btu /lb °F
Ti = 35° F
70 = 2.5 (Tf  35)
Tf 35 = 70/2.5 = 28
Tf = 28 + 35 = 63° F

Just divide both parts of the equation by mC.
Q/mC = mC/mC * (TfTi)
Q/mC = Tf  Ti.
Q/mC + Ti = Tf.
Just put the values in the equation and that's it. If you don't know C for water, it's a "constant" called specific heat. Right now, you just consider C to be a constant value... look it up in google, and you have everything just put it into your calculator and that's it.