Enter the mass of both objects or substances, the initial temperature of each substance, and the specific heat of each substance into the calculator to determine the final temperature of combining the two objects.

## Final Temperature Formula

The following formula is used to calculate the final temperature when combining two substances of different heats.

q1=q2

Where q1 and q2 are the heat of both objects after combination.

Q = m*c*ΔT

TF=(m1*c1*t1+m2*c2*T2)/(m1*c1+m2*c2)

Where TF is the final temperature

## Final Temperature Example Problem

Let’s take a look at a sample problem of how to calculate the final temperature of two combined objects.

For this example, we are going to say we have a hot steel ball that’s dropped into a cooler body of water.

• The initial temperature of the steel is 90C and the initial temperature of the water is 10C.
• The specific heat of the steel is .466 J/gC and the specific heat of water is 4.184 J/gC.
• The mass of the steel is 10g and the mass of the water is 20g.

First, we need to set up the equations:

qwater=qsteel

mwater*Cwater*(Tfinal-Twater)=msteel*Csteel*(Tsteel-Tfinal)

m1*c1*Tf-m1*c1*T1 = m2*c2*T2 – m2*c2*Tf

TF*(m1*c1+m2*c2) = m1*c1*t1+m2*c2*T2

TF=(m1*c1*t1+m2*c2*T2)/(m1*c1+m2*c2)

Next, we enter the values provided above:

20*4.184*( Tfinal – 10 ) = .466*10*( 90 – Tfinal )

83.68* Tfinal – 836.8 = 419.4 – 4.66 * Tfinal

Finally, we re-arrange the equation above, and solve for the final temperature:

88.34* Tfinal = 1256.2

Tfinal = 14.22 C.

In this case, the temperature of the steel dropped significantly and the temperature of the water rose slightly. This is because there was more mass of water than steel and because water has such high specific heat.