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

2 Answers
Aug 29, 2016

Answer:

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.

Sep 4, 2016

Answer:

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"#