Question

An ideal calorimeter is adibiatic, and does not absorb heat from the system it contains. Such a calorimeter contains 250 mL of water at 85°C. To this, a block of ice weighing 25g at -10°C is added. What is the final temperature of the system? Write the appropriate thermodynamic cycle. Assume literature values for all heat capacities, independent of temperature. Values of melting point, ΔHfus, etc. can be taken from literature.

Answer 1

V = 250 ml of water

m = 250 g

T1 = 85°C

m2 = 25 g of ice

Tcie = -10°C

cp ice = 2.03

LAtent heat = 334 J/g

There are going to be the following changes:

Q1 = ice at -10 to ice at 0

Q2 = ice at 0 to water at 0

Q3 = water at 0 to water at TF?

Now, that was for the "cold" substance

for the "hot substance" we have:

Qh = m*Cp*(Tf-Ti) ...

Calculate Q1,Q2

Q1 = m*Cpice*(0--10) = 25*2.03*(0+10) = 507.5 J

Q2 = m*LH = 25*334 = 8350 J

Q3 = 25*4.184*(Tf-0) ....

Now calculate Qh

Qh = m*Cp*(Tf-85) = 250*4.18*(Tf-85)

-Qcold = Qhot

-250*4.18*(Tf-85) = Q1+Q2+Q3

-250*4.18*(Tf-85) =507.5 J+ 8350 J+25*4.184*(Tf-0)

Solve for Tf

-1045(Tf-85) = 8857.5 + 104.5*(Tf)

-1045Tf + 88802 = 8857.5 +104.5Tf

-1045Tf - 104.5Tf = 8857.5 - 88802

-Tf (1149) = -79944.5

Tf = 69.57°C