- #1

kachilous

- 15

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## Homework Statement

A 25 g glass tumbler contains 350 mL of water at 24°C. If four 20 g ice cubes each at a temperature of -3°C are dropped into the tumbler, what is the final temperature of the drink? Neglect thermal conduction between the tumbler and the room.

## Homework Equations

Q = mcT

Q = mL

_{f}(energy required to melt a substance of mass m)

## The Attempt at a Solution

Here are all the known variables:

mass

_{w}= .35kg

c (specific heat for water) = 4186 J/kg * C

T

_{water,initial}= 24 C

mass

_{ice}= 4*20g = .08kg

L

_{f}= 3.33*10

^{5}J/kg

T

_{ice,initial}= -3 C

T

_{f}= ?

First I can use Q = mcT to see if all the ice will get melted by the water

Q = (.35kg)*(4186 J/kgC)*(24C) = 35162.4 J

m = Q/L

_{f}= (35162.4 J) / (3.33*10

^{5}J/kg) = .105592

Since the total mass of the ice is .08, then all the ice will get melted.

Now we can apply conservation of energy to find the final temperature.

mass

_{w}*c*(T

_{water,initial}- T

_{f}) = mass

_{ice}*L

_{f}+ mass

_{ice}*c*(Tf-T

_{ice,initial})

Plugging in all my variables and solving I get 4.18 C which is incorrect.

My math could be wrong but I don't believe it is.

Could anyone provide any insight on what I am doing wrong?

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