Answer : The heat absorbed by the water is 19.4 kJ
Explanation :
Formula used :
[tex]Q=m\times c\times \Delta T[/tex]
or,
[tex]Q=m\times c\times (T_2-T_1)[/tex]
where,
Q = heat absorbed = ?
m = mass of water = 15.5 g
c = specific heat of water = [tex]4.184J/g^oC[/tex]
[tex]T_1[/tex] = initial temperature = [tex]20.0^oC[/tex]
[tex]T_2[/tex] = final temperature = [tex]50.0^oC[/tex]
Now put all the given value in the above formula, we get:
[tex]Q=15.5g\times 4.184J/g^oC\times (50.0-20.0)^oC[/tex]
[tex]Q=1945.56J=1.94\times 10^3J=1.94kJ[/tex]
Therefore, the heat absorbed by the water is 19.4 kJ
