Question

How much heat in calories is needed to raise the temperature of `"125.0 g"` of Lead from `17.5^@"C"` to `41.1^@"C"`? `(c_"Pb"=("0.130 J")/("g"*""^@"C"))`

Answer 1

`"95.6 cal"` are needed.

Explanation:

Use the following equation:

`q=mcDeltaT`,

where:

`q` is heat energy, `m` is mass, `c` is specific heat capacity, and `DeltaT` is the change in temperature. `DeltaT=T_"final"-T_"initial"`

Known

`m="125 g"`

`c_"Pb"=(0.130"J")/("g"*""^@"C")`

`T_"initial"="17.5"^@"C"`

`T_"final"="42.1"^@"C"`

`DeltaT="42.1"^@"C"-"17.5"^@"C"="24.6"^@"C"`

Unknown

`q`

Solution

Plug the known values into the equation and solve.

`q=(125"g")xx((0.130"J")/("g"*""^@"C"))xx(24.6^@"C")="400. J"`
(rounded to three significant figures)

Convert Joules to calories

`"1 J"` `=` `"0.2389 cal"` to four significant figures.

`400. color(red)cancel(color(black)("J"))xx(0.2389"cal")/(1color(red)cancel(color(black)("J")))="95.6 cal"`
(rounded to three significant figures)

`"95.6 cal"` are needed.