Can someone please help me with this? I am following the solutionin the text boo

Question

Can someone please help me with this? I am following the solutionin the text book but I am not completely understanding what it isdoing and when I try new things I get answers that dont make anysense. If you can help me that would be awesome. Thanks.
Here is my problem.
A 25 g glass tumbler contains 390 mL of water at 24°C. If two 15 g ice cubeseach at a temperature of -3°C are dropped into the tumbler,what is the final temperature of the drink? Neglect thermalconduction between the tumbler and the room.


Here is my problem.
A 25 g glass tumbler contains 390 mL of water at 24°C. If two 15 g ice cubeseach at a temperature of -3°C are dropped into the tumbler,what is the final temperature of the drink? Neglect thermalconduction between the tumbler and the room.

A 25 g glass tumbler contains 390 mL of water at 24°C. If two 15 g ice cubeseach at a temperature of -3°C are dropped into the tumbler,what is the final temperature of the drink? Neglect thermalconduction between the tumbler and the room.

Explanation / Answer

Lets convert 390mL=390cm^3 to Kg: density = mass/volume------> (1g/cm^3)(390cm^3)=mass of water                         mass of water = 390g Now, lets say: *Energy necessary to get the water to 0 degrees Celsius :(mass of the water)(change in temperature)(specific heat ofwater)                (0.390)(24)(4186) = 39180.96 Joules *Energy necessary to get ice to 0 degrees Celsius : (mass ofice)(change in temperature)(specific heat of ice)                2(0.015)(3)(2100) = 189 Joules *Energy necessary to completely melt ice : (Latent heat offusion for water)(mass of ice) = (333k)(0.015) = 4995 Joules
From there you can see that there is more than enough energyto completely melt the ice.
Now, Mg = mass of glass Cg = specific hear of glass Mw = mass of water Cw = specific heat of water Mi = mass of ice Ci = specific heat of ice Lf = Latent heat of fusion for water Tf = final temp.
(Mg)(Cg)(change in temp. glass)+(Mw)(Cw)(change in temp.water)=(Mi)(Ci)(0-(-3))+(Lf)(Mi)+(Mi)(Cw)(Tf)
(0.025)(840)(Tf-24)+(0.390)(4186)(Tf-24)=(0.03)(2100)(3)+(333000)(0.03)+(0.015)(4186)(Tf)
Now you can solve for the Final temp. You set up the equation like that because your ice first getsto 0 degrees Celsius (it won't melt if there isn't enough energy).Then, it needs a certain energy to be completely melted (Lf*Mi).Finally, the mass of the ice (which is now water) keeps increasingits temperature because there is still energy left after it wascompletely melted.
Good Luck Now, lets say: *Energy necessary to get the water to 0 degrees Celsius :(mass of the water)(change in temperature)(specific heat ofwater)                (0.390)(24)(4186) = 39180.96 Joules *Energy necessary to get ice to 0 degrees Celsius : (mass ofice)(change in temperature)(specific heat of ice)                2(0.015)(3)(2100) = 189 Joules *Energy necessary to completely melt ice : (Latent heat offusion for water)(mass of ice) = (333k)(0.015) = 4995 Joules
From there you can see that there is more than enough energyto completely melt the ice.
Now, Mg = mass of glass Cg = specific hear of glass Mw = mass of water Cw = specific heat of water Mi = mass of ice Ci = specific heat of ice Lf = Latent heat of fusion for water Tf = final temp.
(Mg)(Cg)(change in temp. glass)+(Mw)(Cw)(change in temp.water)=(Mi)(Ci)(0-(-3))+(Lf)(Mi)+(Mi)(Cw)(Tf)
(0.025)(840)(Tf-24)+(0.390)(4186)(Tf-24)=(0.03)(2100)(3)+(333000)(0.03)+(0.015)(4186)(Tf)
Now you can solve for the Final temp. You set up the equation like that because your ice first getsto 0 degrees Celsius (it won't melt if there isn't enough energy).Then, it needs a certain energy to be completely melted (Lf*Mi).Finally, the mass of the ice (which is now water) keeps increasingits temperature because there is still energy left after it wascompletely melted.
Good Luck

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