Answer:
The right solution is "5.38 grams".
Explanation:
The given values are:
Heat of fusion,
L = 80 cal/g
Mass of ice cube,
[tex]m_{ice} = 41 \ g[/tex]
Specific heat of ice,
[tex]C_{ice}=0.5 \ cal/g[/tex]
Let,
Gram of water freezes will be "m".
⇒ [tex]mL=m_{ice} C_{ice} (0+21)[/tex]
Or,
⇒ [tex]m=\frac{m_{ice} C_{ice} (0+21)}{L}[/tex]
On substituting the values, we get
⇒ [tex]=\frac{41\times 0.5\times 21}{80}[/tex]
⇒ [tex]=\frac{430.5}{80}[/tex]
⇒ [tex]=5.38 \ grams[/tex]
