Thermodynamics problem

Question

Ice of x°C temperature is put into a cylinder of height H=30 сm. Then water (t1=10°C) is poured by the volume of ice. After the thermal equilibrium joint height increased by h=1 сm and became equal 2H+h=61 cm. Find the initial temperature of the ice. Ignore losses.

1. density of ice: 900 kg/m³
2. specific heat of ice: 2100 J/(kg*°C)
3. specific melting heat: 3.3*10⁵ J/kg
4. density of water: 1000 kg/m³
5. specific heat of water 4200 J/(kg*°C)

Please explain this for me. The answer for this problem is -74,6°C

Answer 1

Awesome problem! Just to restate it with some assumptions: A cylindrical tube is filled with 30cm of ice at temp X°C, then filled another 30cm with water at temp 10°C. Once they reach thermal equilibrium (all matter is the same temp), enough of the water has frozen such that the new volume has increased 1cm along the tube.

We can therefore (assuming not all the water has frozen) say that the final temperature is 0°C.

We can volume of a cylinder is the area of the base x height. Therefore, the change in volume can be expressed as

B · h,

where B is the base area and h is the 1cm height change.

We can also describe the change in volume as the volume of the "added" ice - the volume of the "removed" water. Both will have the same mass. So we can therefore say

B · h = Δm/σ_ice - Δm/σ_water = Δm·(1/σ_ice - 1/σ_water)

we can make this in terms of Δm:

Δm = B · h / (1/σ_ice - 1/σ_water)

Now for the thermo part: As it reaches equilibrium, energy is given up from all the liquid water as it cools to 0°C, E_water, as well as by the specific melting heat of some of the water freezing, E_freeze. Meanwhile, all of the original ice warms a total of X°C, absorbing some energy, E_ice.

Since we assume no heat is lost, we can equate these 2 sides:

E_water + E_freeze = E_ice

We can express these using the specific heat values:

(4200J/Kg°C)(10°C)m_water + (3.3x10⁵)Δm = (2100J/Kg°C)(X°C)m_ice

Simplify and solve for X:

X = ( (4.2x10⁴)m_water + (3.3x10⁵)Δm) / (2100)m_ice

We can express the original masses in terms of the original volumes and densities:

m_water = BHσ_water and m_ice = BHσ_ice

If we substitute these and the previous expression for Δm into our expression for X and simplify, we get

X = ( (4.2x10⁴)(0.3)(1000) + (3.3x10⁵)(0.01)/(1/900 - 1/1000)) / (2100)(0.3)(900) = 74.6

And since we've been treating X as the temperature by which the ice rises, its original temp is -74.6°C,