Question

What is the amount of heat required to warm 3.47 moles of water from 25°C to 55°C?

Answer 1

Heat `= 7828.7J`

Explanation:

First convert the moles to grams to use the specific heat of water.
`3.47 " moles" H_2O xx (18g)/"mole" = 62.46g`

Heat (J) `= (4.178 J)/(g*K) xx (62.46g * 30^o) = 7828.7J`

Answer 2

`q="7850 J"`

Explanation:

The molar heat capacity of liquid water is `75.375"J"/("mol" *^@"C")`.
http://www.chemteam.info/Thermochem/Thermochem-Example-Probs5.html

The equation to use is:

`q=n*c_p*DeltaT`,

where:

`q` is the amount of energy gained or lost, `n` is moles, `c_p` is the specific heat capacity, and `DeltaT` is the change in temperature.
`DeltaT=T_"final"-T_"initial"`.

Known

`c_p` `=` `75.375"J"/("mol" *^@"C")`

`n="3.47 mol H"_2"O"`

`DeltaT="55"^@"C"-"25"^@"C"="30"^@"C"`

Unknown

`q`

Solution

Plug in the known values and solve for `q`.

`q=(3.47color(red)cancel(color(black)("mol")))xx(75.375"J"/(color(red)cancel(color(black)("mol")) *^@color(red)cancel(color(black)("C"))))xx(30^@color(red)cancel(color(black)("C")))="7850 J"` (rounded to three significant figures)