Question

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 at – 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.

Answer 1

Step-by-step 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* 2.46+90* 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* 2.108* (8.5)+m* 334`

Comparing 1 and 2 we get

`m=23.65\ gm`

Thus 23.65 gm of ice is added