Question

Calculate the amount of heat transferred when 710 grams of water warms from an initial temperature of 4.0 ºC to a final temperature of 25.0 ºC. The specific heat capacity of liquid water is 4.184 J/g ºC.

Answer 1

Q = 62383.44 Joules

Step-by-step explanation:

Given that,

Mass of water, m = 710 gm

Initial temperature of water,
`T_i=4^(\circ) C`

Final temperature of water,
`T_f=25^(\circ) C`

The specific heat capacity of liquid water is,
`c=4.184\ J/g\ ^oC`

Heat transferred is given by :

`Q=mc(T_f-T_i)`

`Q=710* 4.184* (25-4)`

Q = 62383.44 Joules

So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.