Answer:
A. Final Temp = 36.428C and
47.9g ice will melt
Explanation:
Given the following data:
Mass of water (M1) = 45.0g = 0.045kg
Temperature (T1) = 85C = 358k
Mass of ice (M2) = 105g = 0.105kg
Temperature (T2) = 0c = 273k
Specific heat of water (C) = 4.18j/gC = 0.00418kj/kgc
Molar heat of fusion of water = 6.01kj/mol
Therefore, heat required (q) = MCT
M1C(T1-T2) = M2C∆T
By putting the data we have
0.045×0.00418×(358-273) = 0.105×0.00418×∆T
∆t = 0.045×0.00418×85/0.105×0.00418
∆t = 36.428C
Gram of ice that would melt would be 47.9g
Part(a): The final temperature of the mixture is [tex]36.428^{\circ} C[/tex]
Part(b): The amount of ice that will melt is 57.13 g
Part(a):
Given,
Mass of water = 45.0 g
The initial temperature of water =
Mass of ice = 105.0 g
The temperature of ice =
Molar heat of fusion of water =[tex]6.01 kJ/mol = 0.334 kJ/g[/tex]
Specific heat of water =[tex]4.18 J/gC^{\circ}[/tex]
The formula for the required heat is,
[tex]m_1c(T_1-T_2)=m_2c\bigtriangleup T[/tex]
Now, substituting the given data into the above formula we get,
[tex]\bigtriangleup T= 0.045\times0.00418\times\frac{85}{0.105} \times0.00418\\\bigtriangleup T=36.428^{\circ} C[/tex]
Part(b):
The heat is taken by water to cool from [tex]85^{\circ}C[/tex] to [tex]0^{\circ}C[/tex] is,
[tex]ms\bigtriangleup T=45.0\times4.18\times(85-0)\\= 15988.5 J[/tex]
Heat given by ice to melt at [tex]0^{\circ}[/tex] is,
[tex]105\times0.334=35.07 kJ[/tex]
It may be seen that the amount of ice is in excess to cool water from [tex]85^{\circ}C[/tex] to [tex]0^{\circ}C[/tex] as it gives out more heat than that required by water to cool to [tex]0^{\circ}C[/tex]
Calculation of the amount of ice required to cool water from [tex]85^{\circ}C[/tex] to [tex]0^{\circ}C[/tex] is,
[tex]15.9885 kJ = m \times 0.334 kJ/g\\m=47.87 g[/tex]
So, the ice will melt is,
[tex]105.0 g- 47.87 g= 57.13 g[/tex]
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