Answer:
Incomplete question:
Refer to the temperature versus time graph when answering the questions in Parts C through F. A system consists of 250 of water. The system, originally at = 21.0 , is placed in a freezer, where energy is removed from it in the form of heat at a constant rate. The figure shows how the temperature of the system takes to drop to, after which the water freezes. Once the freezing is complete, the temperature of the resulting ice continues to drop, reaching temperature after an hour. The following specific heat and latent heat values for water may be helpful.
specific heat of ice (at ) = 2.10 J/g K
latent heat of fusion (ice to water phase change at ) = 333.7 J/g
specific heat of water (at ) = 4.186 J/g K
latent heat of vaporization (water to steam phase change at ) = 2256 J/g
specific heat of steam (at ) = 2.01 J/g K
Answers:
Qtotal = 237775 J
Explanation:
To solve this exercise it is necessary to know that if the system is a single phase in which there is a temperature change or if it is a phase change at a single temperature. In the first case, the following formula would be used to calculate the amount of heat:
Q₁ = mCpΔT
Here
m is the mass = 250 g
Cp is the specific heat of ice = 2.1 J/g K
ΔT = 21 - 0 = 21°C = 294 K
In this case the amount of energy is
Q₁ = 250*2.1*294 = 154350 J
In the second case, where there is a phase change at a single temperature, the amount of heat is:
Q₂ = mLf
Here
Lf = latent heat of fusion (ice to water phase change) = 333.7 J/g
Substituting:
Q₂ = 250*333.7 = 83425 J
The total heat is:
Qtotal = 154350+83425=237775 J
