Answer:
Q = 62383.44 Joules
Explanation:
Given that,
Mass of water, m = 710 gm
Initial temperature of water, [tex]T_i=4^{\circ} C[/tex]
Final temperature of water, [tex]T_f=25^{\circ} C[/tex]
The specific heat capacity of liquid water is, [tex]c=4.184\ J/g\ ^oC[/tex]
Heat transferred is given by :
[tex]Q=mc(T_f-T_i)[/tex]
[tex]Q=710\times 4.184\times (25-4)[/tex]
Q = 62383.44 Joules
So, the amount of heat transferred is 62383.44 Joules. Hence, this is the required solution.
