Question

Suppose you add 100 grams of water at 60.0 degrees C to 100

gramsice at 0.00 degrees C. Some of the ice melts and cools thewarm
water to 0.00 degrees C. When the ice/water mixture hascome to a
uniform temperature of 0.00 degrees C, how much ice
hasmelted?

Answer 1

Answer: The amount of ice melted is 50.3 grams.

Step-by-step explanation:

To calculate the heat released by water, we use the equation:

`q=mc\Delta T`

where,

q = heat released

m = mass of water = 100 g

c = specific heat capacity of water = 4.186 J/g.°C

`\Delta T` = change in temperature =
`T_2-T_1=(60-100)^oC=-40^oC`

Putting values in above equation, we get:

`q=100g* 4.186J/g.^oC* (-40)^oC\\\\q=-16744J`

The amount of heat released by water will be absorbed by ice.

So, amount of heat absorbed by ice = -q = -(-16744) J = 16744 J

To calculate the enthalpy change of the reaction, we use the equation:

`\Delta H_(fusion)=(q)/(m)`

where,

`q` = amount of heat absorbed = 16744 J

m = mass of ice melted = ?

`\Delta H_(fusion)` = heat of fusion = 333 J/g

Putting values in above equation, we get:

`333J/g=(16744J)/(m)\\\\m=(16744J)/(333J/g)=50.3g`

Hence, the amount of ice melted will be 50.3 g