Question

Calculate the quantity of heat absorbed by 50 g of water that warms from 30∘C to 90∘C.

Answer 1

Answer:

12,558 J

Explanation:

The amount of heat absorbed by the water is given by:

$Q=mC_S \Delta T$

where

m = 50 g is the mass of the water

$C_s = 4.186 J/gC$ is the water specific heat

$\Delta T=90C-30C=60^{\circ}C$ is the temperature increase of the water

Substituting into the formula, we find

$Q=(50.0 g)(4.186 J/gC)(60^{\circ}C)=12,558 J$