Question

how many joules of heat are released when 110g of water cools from 91 degrees Celsius to 28 degrees celsius

Answer 1

The specific heat of water $c=4.184\frac{J}{g\cdot\degree{C}}$

$Q=cm\Delta{T}=4.184\frac{J}{g\cdot\degree{C}}\times110\ g\times(91\degree{C}-28\degree{C})=2.9\times10^4\ J$

Answer: $2.9\times10^4\ J$