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Lindsey Chheng 1E
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Postby Lindsey Chheng 1E » Tue Feb 11, 2020 3:37 pm

Can someone please help me solve this question?

An ice cube of mass 50.0g at 0.0 Celsius is added to a glass containing 400.0 g of water at 45.0 Celsius. What is the final temperature of the system? Assume no heat is lost to the surroundings.

I know that ice is going through a phase change at 0.0 Celsius and it warms up to some final temperature and that I am supposed to use q(ice) = -q(water)

Thank you!

Posts: 149
Joined: Sat Sep 14, 2019 12:17 am

Re: 4C.13

Postby nicolely2F » Tue Feb 11, 2020 9:23 pm

You know that q(ice) = -q(water). Let's split that into two parts:

A) q(ice) = heat required by phase change + heat required by temperature change
q(ice) = n.∆H + m.c.∆T = (50 / 18).(6.01*10^3) + (50)(4.18)(Tf - 0)
> Here, 6.01*10^3 (the enthalpy of fusion of water per mole) is given in one of the tables of the textbook; (50 / 18) is to obtain the number of moles of ice; and (4.18) is the specific heat capacity of water per gram (which is why in "m.c.∆T" the ice cube's mass remains in grams). Tf is final temperature and 0 is initial temperature

B) -q(water) = energy lost by water
-q(water) = m.c.∆T = (400)(4.18)(Tf - 45)
> Here, 400 is the mass in grams, 4.18 the specific heat capacity, Tf the final temperature, and 45 the initial temperature.

Plug it all in, then use q(ice) = -q(water) to find Tf. Your result should be 31°C.

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