# Specific Heat Capacity Question Mixture Q=MCdeltaT

• dav1d
In summary, the conversation discusses finding the specific heat capacity of an unknown liquid after it is mixed with water. The method involves using the equation Q=m*c*deltaT and assuming that the specific heat capacity of the liquid is similar to that of water. However, this assumption may not hold true in extreme cases such as phase changes. The conversation ends with a discussion about the possible identity of the unknown liquid based on its specific heat capacity.

## Homework Statement

A mixture is made by adding 75g of an unknown liquid at a temperature of 25 degrees Celsius to 60g of water at a temperature of 90 degrees Celsius. The final temperature f the mixture is 65 degrees Celsius. Calculate the specific heat capacity of the liquid. What is the liquid? How are you sure that you are correct?

Q=m*c*deltaT

## The Attempt at a Solution

I don't even know where to start. You can't calculate Q for water because C is not 4.181 at 90 degrees C. You can't find the Q of the unknown liquid because we do not know it's c value...

You are correct that the heat capacity of a substance (C) varies with temperature. However C does not change very much with temperature, so unless we are doing very precise calculations, we usually just assume stays constant at all temperatures.

Ok, so in that case/

Is it correct to find Q for water with c=4.181, delta T = -25 degrees C

set that equal to mcdeltaT for the unknown liquid, rearrange and solve for c of the unknown and then match it up with a table??

That sounds like a good plan.

So c, specific heat capacity, changes only slightly between temperatures, so we can assume they are the same always? When does this not work?

Unless the material changes phase (i.e. liquid water turning into ice or vapor), it's a pretty good assumption that the heat capacity will not change significantly.

See for example this table. Specific heat of liquid water between 0°C and 100°C doesn't vary by more than 1%.

Apparently even without phase change differences can be substantial when temperature approaches critical point.

Ok so
Qwater=mcdeltaT
=60g*4.181J/gdegree C*(65-90degree C)
=-6271.5J

Qunknown=6271.5J (flip signs since the energy is absorbed by the unknown liquid)
c=-6271.5J/(75g)(65-25degree C)
=2.0905J/gdegree C

Now I don't know what it is, according to wikipedia tables for specific heat capacity it should be steam? or is it ice? But the question says liquid...

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Do you have a table with specific heats in your textbook, notes, course materials? If so, look there.

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Borek said:
Do you have a table with specific heats in your textbook, notes, course materials? If so, look there.

I have, but it's pretty much split between ice or steam, which makes no sense since the temperature is too low to form steam and too high to form ice.