Question

`"6345 J"` of energy are required to raise the temperature of water from `22.6^@"C"` to `45.1^@"C"`. What is the mass of the water?

Answer 1

The mass of water is `"67.4 g"`.

Explanation:

Use the equation:

`q=mcDeltat`

where `q` is energy in Joules (J), `m` is mass, `c` is specific heat capacity, and `Deltat`, which is change in temperature. `Deltat=T_"final"-T_"initial"`

The specific heat capacity of water, `c_"H2O"`, is not given. It is `4.184 "J"/("g"*^@"C")`.
https://water.usgs.gov/edu/heat-capacity.html

Organize your information.

Given
`q="6345 J"`
`c_"H2O"=4.184 "J"/("g"*^@"C")`
`Deltat=("45.1"^@"C"-"22.6"^@"C")="22.5"^@"C"`

Unknown: mass in grams

Solution
Rearrange the equation to isolate `m`. Substitute the given values into the equation and solve.

`m=q/(cDeltat)`

`m=(6345color(red)cancel(color(black)("J")))/(4.184color(red)cancel(color(black)("J"))/("g"*^@color(red)cancel(color(black)("C")))xx22.5^@color(red)cancel(color(black)("C")))="67.4 g H"_2"O"` (rounded to three significant figures)