Answer: 135 grams
Explanation:
[tex]Q_{absorbed}=Q_{released}[/tex]
As we know that,
[tex]Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})[/tex]
[tex]m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)][/tex]
where,
[tex]m_1[/tex] = mass of ice = 100 g
[tex]m_2[/tex] = mass of aluminium cup =? g
[tex]T_{final}[/tex] = final temperature =[tex]20^0C[/tex]
[tex]T_1[/tex] = temperature of ice = [tex]-10^oC[/tex]
[tex]T_2[/tex] = temperature of aluminium cup= [tex]70^oC[/tex]
[tex]c_1[/tex] = specific heat of ice= [tex]2.03J/g^0C[/tex]
[tex]c_2[/tex] = specific heat of aluminium cup = [tex] 0.902 J/g^0C[/tex]
Now put all the given values in equation (1), we get
[tex][100\times 2.03\times (20-(-10))]=-[m_2\times 0.902\times (20-70)][/tex]
[tex]m_2=135g[/tex]
Therefore, the mass of the aluminium cup was 135 g.
