How much energy is required to melt a 2.3 kg block of ice? What is the change in entropy of the ice in this process?
In this problem, you are asked for energy required to change the state of matter, which just depends on mass and the tabulated value of latent heat. Entropy is defined as the energy in or out of a system (calculated in part a of the problem) divided by temperature. In each case, you aren’t asked for new information in this problem but rather to restate what you know about the situation in a slightly different way.
There is no need for a picture in most definition problems, and this is one of them. You know (or can look up) melting temperature and latent heat of transformation for ice and are asked for the closely related values of thermal energy and entropy. A picture will not provide any clarity or organization beyond what is already present in the problem.
In equation form, thermal energy needed to change the state of matter is defined as
Q = mL
and change in entropy is defined as
ΔS = ΔQ/T
These are the only relations you need for this problem.
Q = mL
Q = (2.3 kg)(334 kJ/kg)
Q = 770 kJ = 770,000 J
ΔS = ΔQ/T
ΔS = (770,000 J)/(273 K)
ΔS = 2800 J/K
There is no further calculation required in this problem.
Some books use Q to represent the thermal energy in or out of a system, and others use ΔQ. These are all equivalent and refer to the same thing. Likewise, some books use S and others ΔS to refer to the change in entropy during a process.
Latent heats for different substances can be found in your text book or on line. Make sure to use the latent heat of melting and not of vaporization—it takes more energy to vaporize a kg of water than to melt a kg of ice.
Only two significant figures were provided in this problem, so only two figures were kept in the solution.
Whenever temperature appears in a thermal energy equation, you need to work in Kelvin. (The exception is that ΔT is the same in K as it is in ^{0}C because the degree size is the same.) Ice melts at 0 ^{0}C, or 273 K.
Latent heats for different substances can be found in your text book or on line. Make sure to use the latent heat of melting and not of vaporization—it takes more energy to vaporize a kg of water than to melt a kg of ice.
Whenever temperature appears in a thermal energy equation, you need to work in Kelvin. (The exception is that ΔT is the same in K as it is in ^{0}C because the degree size is the same.) Ice melts at 0 ^{0}C, or 273 K.
It takes energy to change the state of matter. During a state change, energy added to the system goes into changing the state. After the change has taken place, energy put into the system raises the temperature. So as long as the system of interest is in thermal equilibrium, temperature remains the same throughout the melting (or boiling) process.
It takes energy to change the state of matter. During a state change, energy added to the system goes into changing the state. After the change has taken place, energy put into the system raises the temperature. So as long as the system of interest is in thermal equilibrium, temperature remains the same throughout the melting (or boiling) process.
Q = mL
Q = (2.3 kg)(334 kJ/kg)
Q = 770 kJ = 770,000 J
ΔS = ΔQ/T
ΔS = (770,000 J)/(273 K)
ΔS = 2800 J/K
This problem is merely a definition problem. You will use the definition of thermal energy in problems that ask, for example, for temperature a change of a system, much as you use F_{g} = mg as you solve force problems.
Note that the sign on entropy matters. In this case, energy was put into the system to melt the ice (Q is positive) and the entropy increased (ΔS is positive.)