Question

If you wish to warm 100 kg of water by 20 degrees C for your bath how much heat is required?

Answer 1

2000000 heat calories.

Explanation:

The heat capacity of water is 1 calorie for 1 gram raised 1 degree C
or 4.186 joules of energy for 1 gram raised 1 degree C ( or `K^o`)

l Kg = 1000 grams so

`100 Kg xx 1000 g /1` = 100000 grams of water.

It takes 10 000 calories to raise 10 000 grams of water 1 degree

so 20 degrees x 10 000 grams = 2 000 000 calories.

The same type calculations can be done for heat in joules

10, 000 grams x 4.186 joules x 20 degrees. = 8372000 joules.

Answer 2

The heat required is 8.4 MJ.

Explanation:

The energy required to heat an object is given by the formula

`color(blue)(|bar(ul(color(white)(a/a) q = mcΔTcolor(white)(a/a)|)))" "`

where

`q` is the energy required
`m` is the mass
`c` is the specific heat capacity
`ΔT` is the change in temperature

In your problem,

`m = "100 kg" = "100 000 g"`
`c = "4.179 J·°C"^"-1"·"g"^"-1"`
`ΔT = "20 °C"`

∴ `q = "100 000" color(red)(cancel(color(black)("g"))) × "4.179 J"·color(red)(cancel(color(black)("°C"^"-1"·"g"^"-1"))) × 20 color(red)(cancel(color(black)("°C"))) = 8.4 ×10^6color(white)(l) "J" = "8.4 MJ"`