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

**Answer 1**

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!

**Answer 2**

Q = mC(Tf-Ti)

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

**Answer 3**

Just divide both parts of the equation by mC.

Q/mC = mC/mC * (Tf-Ti)

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.