Answer:
The Quantity of heat require is [tex]0.062kJ[/tex]
Step-by-step explanation:
This problem bothers on the topic of heat capacity.
Given data
mass of water m= [tex]710 grams[/tex]
converting from grams to kg we have =[tex]\frac{710}{1000} = 0.71kg[/tex]
initial temperature T1= 4°C
final temperature T2= 25°C
heat capacity of water = [tex]4.184 J/g[/tex]
The expression for the quantity of heat required is given has
[tex]Q= mc(T2-T1)[/tex]
substituting our given data we have
[tex]Q= 0.71*4.184*(25-4)\\Q= 2.97*(21)\\Q= 62.38J\\[/tex]
In kilograms we have to divide by [tex]1000[/tex]
[tex]Q= \frac{62.38}{1000} = 0.062kJ[/tex]
