A 52.5 g ice cube, initially at 0°C, is dropped into a Styrofoam cup containing

Question

A 52.5 g ice cube, initially at 0°C, is dropped into a Styrofoam cup containing 366 g of water, initially at 21.0°C. What is the final temperature of the water, if no heat is transferred to the Styrofoam or the surroundings?

Note: I've tried the other methods listed in similar questions here, but I'm not able to get a correct answer. I'm not sure if I'm just making an elementary mistake or if there is something not being taken into account during the explanation that I am assumed to know? If I can trouble anyone kind enough to help me to explain how to reach the solution with the assumption that I'm virtually obtuse, I would greatly appreciate the effort!

Explanation / Answer

given

m1 = 52.5 g

T1 = 0oC

m2 = 366 g

T2 = 21 oC

Let the final temperature is Te.

so,

Te = m1*T1 + m2*T2 / (m1 + m2)

Te = 0.0525*0 + 0.366*21 / (0.0525 + 0.366)

Te = 7.686 / 0.4185

Te = 18.36 oC

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