Question

Suppose that the amount of heat removed when 1.4 kg of water freezes at 0.0 oC were removed from ethyl alcohol at its freezing/melting point of -114.4 oC. How many kilograms of ethyl alcohol would freeze?

Answer 1

`m = 4.26 kg`

Step-by-step explanation:

As we know that heat required to remove from water to freeze it is given as

`Q = mL`

so we have 1.4 kg water is required to freeze

so heat is given as

`Q_1 = 1.4 * 3.35 * 10^5`

now same heat is removed from ethyl alcohol of mass "m"

the latent heat of fusion for ethyl alcohol is given as

`L = 1.1 * 10^5 J/kg`

so we have

`Q_2 = m(1.1 * 10^5)`

now we know

`Q_1 = Q_2`

`1.4 * 3.35 * 10^5 = m(1.1 * 10^5)`

`m = 4.26 kg`