Answer : The heat required is, 1350 J
Solution :
Formula used :
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
where,
Q = heat required = ?
m = mass of aluminum = 30 g
c = specific heat of aluminum = [tex]0.900J/g^oC[/tex]
[tex]\Delta T=\text{Change in temperature}[/tex]
[tex]T_{final}[/tex] = final temperature = [tex]75^oC[/tex]
[tex]T_{initial}[/tex] = initial temperature = [tex]25^oC[/tex]
Now put all the given values in the above formula, we get the heat required.
[tex]Q=30g\times 0.900J/g^oC\times (75-25)^oC[/tex]
[tex]Q=1350J[/tex]
Therefore, the heat required is, 1350 J
