Question

A 49 lb piece of steel at 670 F is dropped into 95 lbs of water at 50 F. What is the final temperature of the mixture.

Answer 1

`83.82F`

Step-by-step explanation

Parameters given:

Mass of steel, M = 49 lb = 22.23 kg

Initial temperature of steel, ts = 670 F = 627.59 K

Mass of water, m = 95 lb = 43.09 kg

Initial temperature of water, tw = 50 F = 283.15 K

We want to find the final temperature of the mixture of the piece of steel and water.

The amount of heat released in the system must be equal to the total amount of heat absorbed by the system:

`Q_R=Q_A`

The piece of steel releases heat while the water absorbs the heat. This implies that the equation above becomes:

`-MC(T_s-t_s)=mc(T_w-t_w)_{}`

where C = specific heat capacity of steel = 468 J/kgK

c = specific heat capacity of water = 4184 J/kgK

Ts = final temperature of steel

Tw = final temperature of water

Since the final temperatures of both the steel and water are equal, it implies that:

`T_s=T_w=T`

Substituting the given values into the equation above, we have that:

`-22.23\cdot468\cdot(T-627.59)=43.09\cdot4184\cdot(T-283.15)`

Simplify the equation above:

`\begin{gathered} -10403.64(T-627.59)=180288.56(T-283.15) \\ \Rightarrow-10403.64T+6529220.428=180288.56T-51048705.76 \\ \Rightarrow180288.56T+10403.64T=6529220.428+51048705.76 \\ \Rightarrow190692.2T=57577926.19 \\ T=(57577926.19)/(190692.2) \\ T=301.94K \end{gathered}`

Let us now convert to Fahrenheit:

`T=((301.94-273.15)\cdot9)/(5)+32=83.82F`

That is the final temperature of the mixture.