Question

what mass of water at 15 degrees celcius can be cooled 1 degree celcius by heat necessary to melt 185 grams of ice at 0 degrees celcius?

Answer 1

The required mass is 14.741 kg,

Explanation:

Consider the provided information.

Heat of fusion of ice
`h_f =333.55J/g`

Heat necessary to melt 185 grams of ice at 0 degrees Celsius is:

`Q_(ice)=m_(ice)\cdot h_f =185*333.55 =61706.75 J`

The heat gained by water is:
`Q_w=-Q_(ice)=-61706.75 J`

Specific heat capacity of water is
`C_W=4.186 J/g`°C

Now use the formula:
`Q_W=m_w\cdot C_w\cdot dT`

Substituting the respective values in the above formula.

`-61706.75=m_w\cdot 4.186(0-1)`

`-61706.75=m_w\cdot (-4.186)`

`m_w=(-61706.75)/(-4.186)\\m_w=14741.22g\ or\ 14.741 kg`

Hence, the required mass is 14.741 kg,