Question

If 200 grams of water is to be heated from 24°C to 100°C to make a cup of tea, how much heat must be added?

The specific heat of water is 4.18 `J` `/` `g*C`.

Answer 1

`q = "64 kJ"`

Explanation:

As you know, a substance's specific heat tells you how much heat is required to increase the temperature of `"1 g"` of that sample by `1^@"C"`.

In water's case, you know that its specific heat is equal to `4.18"J"/("g" ""^@"C")`. This tells you that in order to increase the temperature of `"1 g"` of water by `1^@"C"`, you need to supply `"4.18 J"` of heat.

How much heat would you need to increase the temperature of `"200 g"` of water by `1^@"C"`?

Well, if you need `"4.18 J"` per gram to increase its temperature by `1^@"C"`, it follows that you will need `200` times more heat to get this done.

Likewise, if you were to increase the temperature of `"1 g"` of water by `76^@"C"`, you'd need `76` times more heat than when increasing the temperature of `"1 g"` by `1^@"C"`.

Combine these two requirements and you get the total amount of heat required to increase the temperature of `"200 g"` of water by `76^@"C"`.

Mathematically, this is expressed using the following equation

`color(blue)(q = m * c * DeltaT)" "`, where

`q` - heat absorbed/lost
`m` - the mass of the sample
`c` - the specific heat of the substance
`DeltaT` - the change in temperature, defined as final temperature minus initial temperature

Plug in your values to get

`q = 200 color(red)(cancel(color(black)("g"))) * 4.18"J"/(color(red)(cancel(color(black)("g"))) color(red)(cancel(color(black)(""^@"C")))) * (100 - 24)color(red)(cancel(color(black)(""^@"C")))`

`q = "63536 J"`

I'll leave the answer rounded to two sig figs and expressed in kilojoules

`q = color(green)("64 kJ")`