Question

How much energy will be required to heat a `"1.0 kg"` mass of water from `25^@"C"` to `99^@"C"`?

Answer 1

You'd need `"310 kJ"` to raise the temperature of 1 kg of water from `25` to `99^@"C"`.

So, you have all the information you need to use the equation

`q = m * c * DeltaT`, where

`q` - the amount of heat needed;
`m` - the mass of water - in your case 1.0 kg:
`c` - the specific heat of water;
`DeltaT` - the difference between the final temperature, `99^@"C"`, and the initial temperature, `25^@"C"`, of the water.

Plug your data into the equation to get

`q = "1,000 g" * 4.18"J"/("g" * ^@"C") * (99-25)^@"C"`

`q = "309,320 J" = "+309.3 kJ"`

Rounded to two sig figs, the answer will be

`q = "+310 kJ"`

Answer 2

It will require `"310,000 J"`.

Explanation:

In order to answer this question, you you will need to use the following equation:

`q = cmDeltaT`,

where `q` is the quantity of heat gained or lost, `c` is the specific heat capacity (of water in this case), `m` is mass in grams, and `DeltaT` is the difference in temperature, `DeltaT=T_"final"-T_"initial"`

Known/Given:
`c_"water"= 4.184 "J"/("g"*""^("o")"C"`

`m=1.0color(red)cancel(color(black)("kg"))xx"1000 g"/(1color(red)cancel(color(black)("kg")))=1.0xx10^3` `"g"`

`T_i="25"^@"C"`

`T_f="99"^@"C"`

`DeltaT="99"^@"C" - "25"^@"C"="74"^@"C"`

Unknown:
`q`

Solution:

Plug the known values into the equation and solve.

`q=4.184"J"/(color(red)cancel(color(black)("g"))*""^@color(red)cancel(color(black)("C")))xx1.0xx10^3color(red)cancel(color(black)("g"))xx74^@color(red)cancel(color(black)("C")) = "310,000 J"` (rounded to two significant figures)