A 325 gram piece of gold at 427 C is dropped into 200mL of water at 22.0 C. The specific heat of gold is 0.031 cal/g- C. Calculate the final temperature of the mixture. ( assume no heat losses to the surroundings and take care to use the proper signs, + or -)

325 gram of gold at 427 C
200mL of water at 22.0 C

The specific heat of gold is 0.031 cal/g-C
The specific heat of water is 1.0 cal/g-C
The density of water is 1.0 g/mL

Assume that no phase changes occur.


325\,g \,\times\, 0.031 \frac {cal}{gC} = 10.075 \,\frac {cal}{C}

200\,g \,\times\, 1.0 \frac {cal}{gC} = 200 \,\frac {cal}{C}

T = Final Temperature
Caloric decrease of gold will equal caloric increase of water.

10.075 \,\times\, (427-T) = 200 \,\times\, (T-22)

Solve for T.

In case you're wondering why Perito did this, we make use of the formula Q=mcT

The heat involved is the same in both cases, so, Q_{gold} =Q_{water} and (mcT)_{gold} = (mcT)_{water}.

So, 325\, \times 0.031\, \times (427-T) = 200\, \times\, 1.0 \,\times(T-22) .