Question

If 75 g of liquid water (C=4.18 J/g- degree C) in a calorimeter changes temperature from 25 degree C to 75 degree C how much heat was transferred

Answer 1

Answer : 15675 J of heat was transferred.

Solution : Given,

Mass of liquid water = 75 g

Specific heat capacity =
`4.18J/g^oC`

Initial temperature =
`25^oC`

Final temperature =
`75^oC`

Formula used :

`q=mc\Delta T\\\\q=mc(T_(final)-T_(initial))`

where,

q = heat energy

m = mass

c = specific heat capacity

`\Delta T` = change in temperature

`T_(initial)` = initial temperature

`T_(final)` = final temperature

Now put all the given values in the above formula, we get

`q=mc(T_(final)-T_(initial))`

`q=(75g)* (4.18J/g^oC)* (75-25)^oC`

`q=15675J`

Therefore, 15675 J of heat was transferred.