Question

when 200 grams of water cools from 50 c to 25 c the total amount of heat energy released by the water is?

Answer 1

Specific Heat:
Heat Energy= Mass of substance X Specific Heat X Change in Temp.
1. change in temp |50-25| = 25
2. specific heat of Water(H2O) = cal/g (Celsius) 1.000
heat energy= 200g X 1.000 X 25
Heat energy = 5000cal

Answer 2

Answer : The heat released by the eater is,
`2.1* 10^4J`[/tex]

Explanation :

Formula used :

`Q=m* c* \Delta T`

or,

`Q=m* c* (T_2-T_1)`

where,

Q = heat released = ?

m = mass of water = 200 g

c = specific heat of water =
`4.184J/g^oC`

`T_1` = initial temperature =
`50^oC`

`T_2` = final temperature =
`25^oC`

Now put all the given value in the above formula, we get:

`Q=200g* 4.184J/g^oC* (25-50)^oC`

`Q=20920J=2.1* 10^4J`[/tex]

Therefore, the heat released by the eater is,
`2.1* 10^4J`[/tex]