Question

What was the heat lost from the system if `"5.25 g"` of water was cooled from `64.8^@ "C"` to `5.5^@ "C"`? `C_P = "4.184 J/g"^@ "C"`

Answer 1

`Q_"lost"=ul?J`

Explanation:

Set up the following processes to solve the problem with ease:

`GIVEN:`

`m=5.25g`, the mass of water

`C_p=(4.186J)/(g*^oC)`, specific heat capacity of `H_2O_(liquid)`; http://hyperphysics.phy-astr.gsu.edu/hbase/Tables/sphtt.html

`T_f=5.5^oC`, the initial temperature

`T_i=64.8^oC`, the final temperature

`REQUIRED:`

`Q_"lost"=?`

This is a negative value , as heat moves out from the system.

`SOLUTION:`

The heat flow is found based on the mass involved, its heat capacity, and the changes of temperature in the process; thus,

`Q=mC_pDeltaT`

where:

`DeltaT=T_f-T_i`

Now, plug in identified value to each variable in the formula as shown below. Make sure the units work out and the desired unit as required is attained :

`Q=mC_p(T_f-T_i)`

`Q=5.25cancel(g)xx(4.186J)/cancel(g*^@C)xx(5.5-64.8)cancel(*^@C)`
`Q=color(red)(-)ul ?J` or

`Q_"lost"=ul?J`

Note:

The negative sign`color(red)(-)` here indicates heat lost in the process.