# Using formula Q= mC(Tf-Ti)…. I don’t understand this?

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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• 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(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

• 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.