# How many calories are absorbed by a pot of water with a mass of 500 g in order to raise the temperature from 20° C to 30° C?

Jan 15, 2016

The water absorbs 5000 cal.

#### Explanation:

I assume that the pot itself absorbs no heat.

The formula for the heat absorbed by the water is

q = mcΔt, where

$m = \text{500 g}$
$c = 1.00 \textcolor{w h i t e}{l} {\text{cal°C"^(-1)"g}}^{- 1}$
ΔT = T_2 –T_1 = "30 °C – 20°C" = "10 °C"

q = mcΔT = 500 color(red)(cancel(color(black)("g"))) ×1.00color(white)(l) "cal"color(red)(cancel(color(black)("°C"^(-1)"g"^(-1)))) × 10 color(red)(cancel(color(black)("°C"))) = "5000 cal"