Check out some similar questions!

ezzye · Sep 16, 2009, 06:54 PM

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 -)

Perito · Sep 17, 2009, 06:56 AM

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.

Unknown008 · Sep 17, 2009, 07:50 AM

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