This problem was given to us as a hypothetical in lab: A calorimeter cup whose w

Question

This problem was given to us as a hypothetical in lab:

A calorimeter cup whose water equivalent is WE_c=8.0g contants 250.0g of water at an initial temp of 22.0ºC. If 45.0 grams of ice at 0ºC is added, describe (quantitiatevly) the resulting mixture.

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In the experimental part of the lab, we calculated the latent heat of vaporization and fusion, so I understand how to do that. However, I don't know what "water equivalent" means, and every definition I find on the web just confuses me more. Can someone explain to me in a common sense way what WE is, and how to solve this problem? Thank you so much.

Explanation / Answer

The mass of the water m = 250 + 8 = 0.258kg

the initial temperature Ti = 22 C

The mass of the ice mice = 0.045kg

From calorimetry

   the amount of heat lost by water and water equivalent = heat gained by ice

    mcT = mice L + mice cT

(0.258 kg)(4186)( 22 - T) = (0.045kg)(333*10^3 J/kg) + (0.045)(4186)T

23759.736 - 1079.988T= 14985 + 188.37 T

1268.358T = 8774.74

Then the final temperature T = 6.9 deg

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