Question

A 41 g Ice cube at -21 C is dropped into a container of water at 0 C. How much water freezes onto the ice? The specific heat of ice is .5 cal/g C and it's heat of fusion of is 80 cal/g.

Answer 1

The right solution is "5.38 grams".

Step-by-step explanation:

The given values are:

Heat of fusion,

L = 80 cal/g

Mass of ice cube,

`m_(ice) = 41 \ g`

Specific heat of ice,

`C_(ice)=0.5 \ cal/g`

Let,

Gram of water freezes will be "m".

⇒
`mL=m_(ice) C_(ice) (0+21)`

Or,

⇒
`m=(m_(ice) C_(ice) (0+21))/(L)`

On substituting the values, we get

⇒
`=(41* 0.5* 21)/(80)`

⇒
`=(430.5)/(80)`

⇒
`=5.38 \ grams`