Physics Heat Help

Question

1.) How much heat is necessary to boil away a liter of water beginning at 20°C.

2.) A small pond used by cattle for drinking water is covered by a layer of ice with a mass of 200 kilograms. How much heat from the sun will be required to melt the ice completely?

Answer 1

Hello!

Here is a quick walkthrough of your two questions, please don't hesitate to let me know if you have any other questions afterwards!

Question #1

How much heat is necessary to boil away a liter of water beginning at 20°C.

To "boil away" this liter of water into steam vapor I must do two things: (1) Heat up the water from 20°C to its boiling temperature of 100°C, and then I must (2) supply extra energy required to change the liter of water from liquid phase into vapor phase.

Let's start by calculating the heat required for process (1)...

The thermodynamic equation that describes the quantity of heat/energy required to change the temperature of a substance is written as:

ΔQ = c * m * ΔT

where ΔQ is the heat/change-in-energy required, c is a material-specific property called the "specific heat", m is the mass of material in question, and ΔT is the change in temperature being imposed.

For water, we can look up its specific heat: c = 4184 J / kgK, meaning that it takes 4.184 Joules of energy to heat up 1 kilogram of water by 1 degree Kelvin/Celsius.

For the mass, we actually aren't given the mass of water, but instead are given its volume of 1 L. As such we need to use the density of water to find out how much 1 liter of water weighs. Fortunately, water's density is SUPER convenient: ρ_water = 1 kg / L, so that 1 liter of water has a mass of exactly 1 kg.

For the change in temperature, we can calculate is as ΔT = T_final - T_initial = (100°C) - (20°C) = 80°C

Note: for CHANGES in temperature terms, it doesn't matter if we plug in Celsius or Kelvin.

Now let's plug in:

ΔQ = c * m * ΔT = (4184 J / kgK) * (1 kg) * (80°C) = 334,720 J = 334.72 kJ

So we know that it takes 334.72 kJ to heat our water up to its boiling temperature -- but it will take some extra energy to actually change to vapor phase!

That brings us to process (2)...

Whenever we change the phase of a substance from a more condensed state (like a liquid) to a less condensed state (like a vapor) there is an energetic cost! For boiling a liquid into a vapor, we call this energetic cost the heat of vaporization.

This is again a material specific parameter, so we can look it up for water to be equal to 2,260 kJ / kg.

This means that, at the boiling temperature, it "costs" 2,260 kJ of heat energy in order to convert 1 kg of liquid water into vapor water (steam). In general, we can find the total energy required to convert an arbitrary quantity of material by multiplying this heat of vaporization by our mass.

Since we happen to have exactly 1 kg of liquid water in this problem, the heat energy required to change it into a vapor phase would just be (2,260 kJ / kg) * (1 kg) = 2,260 kJ

So, to wrap up this problem, it takes 334.72 kJ of heat to warm the water up to its boiling temperature, and then it takes an extra 2,260 kJ of heat to then convert it into a vapor. Thus it takes a total heat input of (334.72 kJ) + (2,260 kJ) = 2594.72 kJ of heat to perform the entire process.

Question #2

A small pond used by cattle for drinking water is covered by a layer of ice with a mass of 200 kilograms. How much heat from the sun will be required to melt the ice completely?

At first glance this seems categorically similar to the previous problem, but there is some secret trickiness here!

In case you don't see the trickiness -- What is the initial temperature of the ice at the start of this problem?

In the last problem we were told that the water started at 20 degrees Celsius and that was vital information to solving the problem... At the start is the ice in this problem at 0ºC? -10ºC? -50ºC? -100ºC? How can we know? Well we CAN know, its just tricky!

We actually KNOW that the initial temperature of the ice at the start of the problem is 0ºC because the ice is on top of a lake of liquid water. Whenever we have two phases of a material in thermodynamic equilibrium with each other, we know that their temperature is exactly equal to their phase transformation temperature.

For example, a glass containing an equilibrium mixture of liquid water with solid ice should have a temperature of 0ºC, and a closed cylinder containing an equilibrium mixture of liquid water and vapor water steam should have a temperature of 100ºC.

Since the ice in this problem is ALREADY at its melting temperature (0ºC) at the start of the problem, we don't need to supply any heat to raise its temperature to that point -- we only need to supply energy to change its phase.

Like the previous problem, we are analyzing a thermodynamic process which involves a phase transformation from a more condensed phase (solid ice) into a less condensed phase (liquid water). But we won't be able to use the same heat of vaporization number from earlier, since we are melting into a liquid not vaporizing into a gas. No problem though! We can look up the heat of melting/fusion of water to be 334 kJ/kg ("it takes 334 kJ of heat to melt 1 kg of ice at its melting temperature").

Now, since we know that there is 200 kilograms of ice present in this problem, we can calculate the total heat of melting required for the phase change as (334 kJ/kg) * (200 kg) = 66,800 kJ.

Hope this helps! Let me know if you have any other questions and I'd be happy to help!

--Zach