Question

A 600g copper dish contains 1500g of water at 20 0 C. A 100g iron bar at
120 0 C is dropped into the water. What is the final temperature of the water?

Answer 1

Answer:

The heat added to the water must equal that lost by the iron bar. Delta H iron = C specific × mass of iron × delta T. and similarly for water. Look up the specific heat capacities of the two substances. Iron = 0.45 J/ g °C. Water = 4.184 J/g °C .

Since you don't specify the material from which the dish is made we cannot estimate how much it will be heated! So we will pretend that it isn't affected by the heat added from the iron bar. You also don't specify if it's a closed or open system so we will pretend it's in a piston cylinder which doesn't absorb heat . Do you understand why these assumptions are necessary?

Then since Delta T = Tf - Ti. ( final T - initial T ) we can write…

Delta Hiron = Ciron × mass iron × ( Tf - Ti ) and

Delta H water = Cwater × mass water × ( Tf - Ti )

Since DT water = - DT iron then DH water = - DH iron ( I'm using D to represent Delta)

C w × m w × ( Tf - Ti ) = - ( C ir × m ir × ( Tf - Ti )) = Cir × mir × ( Ti-Tf)

C w × m w × Tf - C w × m w Ti = C ir × m ir × Ti - C ir × m ir × Tf

Let's put in the numbers…

4.184 × 1500 × Tf - 4.184 × 1500 × 20 = 0.45×100×120 - 0.45× 100×Tf

6276×Tf - 125520 = 5400 - 45 ×Tf

6276Tf + 45Tf = 5400 + 125520 = 130920

Tf ( 6276 +45 ) = 130920

Tf = 130920/ 6321 = 20.71 °C not unexpectedly a small increase

I hope that's all clear.

If we knew the heat capacity of the container ( Ti = 20 °C of course) the final temperature would be even closer to 20 °C.