Answer : 15675 J of heat was transferred.
Solution : Given,
Mass of liquid water = 75 g
Specific heat capacity = [tex]4.18J/g^oC[/tex]
Initial temperature = [tex]25^oC[/tex]
Final temperature = [tex]75^oC[/tex]
Formula used :
[tex]q=mc\Delta T\\\\q=mc(T_{final}-T_{initial})[/tex]
where,
q = heat energy
m = mass
c = specific heat capacity
[tex]\Delta T[/tex] = change in temperature
[tex]T_{initial}[/tex] = initial temperature
[tex]T_{final}[/tex] = final temperature
Now put all the given values in the above formula, we get
[tex]q=mc(T_{final}-T_{initial})[/tex]
[tex]q=(75g)\times (4.18J/g^oC)\times (75-25)^oC[/tex]
[tex]q=15675J[/tex]
Therefore, 15675 J of heat was transferred.
