A 500 mL bottle of spring water, which is at room temp. of25 o C, is poured over

Question

A 500 mL bottle of spring water, which is at room temp. of25oC, is poured over 120g of ice that is at-8oC. What will be the final temperature of the waterwhen all the ice has melted, assuming that it is in an insulatedcontainer that does not change temperature? Thanks for any help on this guys.. I know it's simple, but theproblem is really confusing me.. A 500 mL bottle of spring water, which is at room temp. of25oC, is poured over 120g of ice that is at-8oC. What will be the final temperature of the waterwhen all the ice has melted, assuming that it is in an insulatedcontainer that does not change temperature? Thanks for any help on this guys.. I know it's simple, but theproblem is really confusing me..

Explanation / Answer

Heat absorbed by spring water =                     Heat to bring ice to 0 C + Heat released by ice to form liquid +Heat to bring temperature of liquid to final T Heat absorbed by spring water = 500 mL *(1g/mL)*(4.18 J/g/C)*(25 C- Tfinal) Heat to bring ice to 0C from -8 C = 120 g*(2.05 J/g/C)*(0 C - (-8C)) = 1968 J     (ice has heat capacity of 2.05J/g/C) Heat to form liquid from ice = 120g *334 J/g = 40080J                   (enthalpy to melt ice is 334 J/g) Heat to bring temperature of liquid to final T = 120g* (4.18 J/g/C)*(Tfinal - 0 C) So 500 mL *(1g/mL)*(4.18 J/g/C)*(25 C - Tfinal) = 1968 J + 40080 J +120g* (4.18 J/g/C) *(Tfinal - 0 C) 2090*(25 - Tfinal) = 42048 + 501.6*(Tfinal-0) 52250 - 2090*Tfinal = 42058 + 501.6*Tfinal 52250 - 42058 = Tfinal*(501.6 + 2090) Tfinal = (52250 - 42058)/(501.6 + 2090) Tfinal = 3.93 Celsius

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