Question

If a piece of silver specific heat .2165 j/g °C with a mass of 14.16 g and a temperature of 133.5°C is dropped into 250.0 g of fat 17.20°C what will be the final temperature of the system

Answer 1

`T_F=17.56\°C`

Step-by-step explanation:

Hello there!

In this case, for this calorimetry problem, it is possible for us to realize that the heat lost by the hot silver is gained by the cold far whose specific heat is 3.94 J/g°c, so we can write:

`-Q_(Ag)=Q_(fat)`

Which can be written in terms of mass, specific heat and temperature as shown below:

`-m_(Ag)C_(Ag)(T_F-T_(Ag))=m_(fat)C_(fat)(T_F-T_(fat))`

In such a way, solving for the final temperature, we obtain:

`T_F=(m_(Ag)C_(Ag)T_(Ag)+m_(fat)C_(fat)T_(fat))/(m_(Ag)C_(Ag)+m_(fat)C_(fat))}`

Then, we plug in the given data to obtain:

`T_F=(14.16g*0.2165J/g\°C*133.5\°C+250g*3.94J/g\°C*17.20\°C)/(14.16g*0.2165J/g\°C+250g*3.94J/g\°C) \\\\T_F=17.56\°C`

Best regards!