# Discussion on BITSAT

Posted on: 02:37:20 PM on October 11, 2016
16g of oxygen at 37°C is mixed with 14 g of nitrogen at 27°C, the temperature of the mixture will be?

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## Reponses To This

Posted on: 09:24:12 AM on October 13, 2016
Hi Sanu!

Given:
Oxygen: 16g (37°C)
Nitrogen: 14g (27°C)

They aren't equal in mass, but they are equal in terms of moles.
Both are diatonic elements meaning O2 and N2.
A mole of O2 has a mass of 32 g. The problem states the we have 16 grams. This means there is 0.5 moles of oxygen.
1 mole of N2, has a mass of 28 g. If there 14 grams present then there is 0.5 moles of nitrogen.

So the quantities are equal, and the temperatures will average out.
If you have 50 ml of water at 30 degrees C and 50 ml of water at 40 degrees C,
the resulting mixture would have a temperature of 35 degrees. Similar concept.

Posted on: 12:41:11 PM on December 27, 2017
Can u plz give answer in terms of proper steps

Posted on: 11:11:47 AM on February 15, 2019
He is telling to take average of temperature 37+27/2=32

Posted on: 09:08:29 AM on March 7, 2019
Use formula: Final Temperature= (n1T1+ n2T2)/ n1+n2 Hence we will get the ans as 32℃ 