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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.

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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