Question

How much heat, in joules, is absorbed by 500.0 g of water as its temperature is increased from 10.0oC to 85.0oC?  The specific heat of water is 4.184 J/g•oC 

Answer 1

We know the formula,

$Q= nC_v \Delta T$

$Q = \dfrac{500}{18} \times 4.184 \times (85-10)$

$Q = 8716.66J$