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

Answer:

The required mass is 14.741 kg,

Step-by-step 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:

$Q_{ice}=m_{ice}\cdot h_f =185\times333.55 =61706.75 J$

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.

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

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

$m_w=\frac{-61706.75}{-4.186}\\m_w=14741.22g\ or\ 14.741 kg$

Hence, the required mass is 14.741 kg,