Answer:
[tex]m_{leftover}=37g[/tex]
Explanation:
Hello!
In this case, since the melting process involves the heat added to the system, the heat of fusion and the mass of water:
[tex]Q=m\Delta H_{fus}[/tex]
Thus, as it has been reported that the heat of fusion of ice is 333.6 J/g, we can compute the melted mass of ice as shown below:
[tex]m=\frac{Q}{\Delta H_{fus}} =\frac{46000J}{333.6J/g}\\\\m=138g[/tex]
Thus, the ice leftover results:
[tex]m_{leftover}=175g-138g\\\\m_{leftover}=37g[/tex]
Regards!
