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WITCHER [35]
3 years ago
8

Suppose 310. grams of ethanol (ethyl alcohol) is in an aluminum cup of 90.0 grams. Both of these are at 30.0C. A mass m of ice a

t – 8.5C is taken from a freezer and added to the alcohol in the cup. The final temperature of all the components is 18.0C. Assuming no heat was lost from the system, calculate the mass m of the ice added.
Physics
1 answer:
Ira Lisetskai [31]3 years ago
8 0

Answer:

Explanation:

Given

mass of ethanol m_e=310\ gm

mass of aluminium cup m_{al}=90\ gm

both are at an initial temperature of T_i=30^{\circ}C

specific heat of ethanol c_e=2.46\ J/g-K

specific heat of aluminium c_{al}=0.9\ J/g-K

specific heat of ice c_i=2.108\ J/g-K

specific heat of water c_w=4.184\ J/g-K

Latent heat of fusion L=334\ J/gm

suppose m is the mass of ice added

Heat loss by Al cup and ethanol after 18^{\circ}C is reached

Q_1=(310\times 2.46+90\times 0.9)\cdot (30-18)

Heat gained by ice such that ice is melted and reached a temperature of 18^{\circ}C

Q_2=m\times 2.108\times (8.5)+m\times 334

Comparing 1 and 2 we get

m=23.65\ gm

Thus 23.65 gm of ice is added

                 

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