Answer:
Explanation:
Given
mass of ethanol [tex]m_e=310\ gm[/tex]
mass of aluminium cup [tex]m_{al}=90\ gm[/tex]
both are at an initial temperature of [tex]T_i=30^{\circ}C[/tex]
specific heat of ethanol [tex]c_e=2.46\ J/g-K[/tex]
specific heat of aluminium [tex]c_{al}=0.9\ J/g-K[/tex]
specific heat of ice [tex]c_i=2.108\ J/g-K[/tex]
specific heat of water [tex]c_w=4.184\ J/g-K[/tex]
Latent heat of fusion [tex]L=334\ J/gm[/tex]
suppose m is the mass of ice added
Heat loss by Al cup and ethanol after [tex]18^{\circ}C[/tex] is reached
[tex]Q_1=(310\times 2.46+90\times 0.9)\cdot (30-18)[/tex]
Heat gained by ice such that ice is melted and reached a temperature of [tex]18^{\circ}C[/tex]
[tex]Q_2=m\times 2.108\times (8.5)+m\times 334[/tex]
Comparing 1 and 2 we get
[tex]m=23.65\ gm[/tex]
Thus 23.65 gm of ice is added
