Answer: The amount of ice melted is 50.3 grams.
Explanation:
To calculate the heat released by water, we use the equation:
[tex]q=mc\Delta T[/tex]
where,
q = heat released
m = mass of water = 100 g
c = specific heat capacity of water = 4.186 J/g.°C
[tex]\Delta T[/tex] = change in temperature = [tex]T_2-T_1=(60-100)^oC=-40^oC[/tex]
Putting values in above equation, we get:
[tex]q=100g\times 4.186J/g.^oC\times (-40)^oC\\\\q=-16744J[/tex]
The amount of heat released by water will be absorbed by ice.
So, amount of heat absorbed by ice = -q = -(-16744) J = 16744 J
To calculate the enthalpy change of the reaction, we use the equation:
[tex]\Delta H_{fusion}=\frac{q}{m}[/tex]
where,
[tex]q[/tex] = amount of heat absorbed = 16744 J
m = mass of ice melted = ?
[tex]\Delta H_{fusion}[/tex] = heat of fusion = 333 J/g
Putting values in above equation, we get:
[tex]333J/g=\frac{16744J}{m}\\\\m=\frac{16744J}{333J/g}=50.3g[/tex]
Hence, the amount of ice melted will be 50.3 g
