Question

How much thermal energy is

Answer 1

Answer:

Qt = 940500 [J]

Explanation:

In order to solve this problem and understand it well, we must analyze that the ice melts or changes phase at a temperature of 0 [°c], then the water at a temperature of 100 [°C] evaporates or changes phase. That is, we have 5 different stages of water (initially in the form of ice), which are:

-) From -30 [°C] to 0 [°C] = Sensible change of temperature

-) At 0 [°C] = Latent heat of melting ice = 334 x 10³ [J/kg]

-) From 0 [°C] to 100 [°C] = Sensible change of temperature

-) At 100 [°C] = Latent heat of fusion for water = 2260 x 10³ [J/kg]

-) From 100 [°C] to 130 [°C] = Sensible change of temperature (superheating)

We can calculate the sensible heat of any process with the following expression:

$Q=m*C_{p}*(T_{f}-T_{i})$

where:

Q = heat or thermal energy [J]

m = mass of the body or substance = 300 [g]

Cpw = specific heat of the water = 4180 [J/kg*°C]

Cpi = specific heat of the ice = 2090 [J/kg*°C]

Tfinal = final temperature of the process [°C]

Tinicial = initial temperature of the process [°C]

And for some latent heat process, we can calculate the thermal energy using the following expression:

$Q_{lat}=m*Dh_{fus/eva}$

where:

Dhfus/eva = Fusion heat or latent heat [J/kg]

So, we need to calculate the fice processes.

$Q_{1}=0.3*2090*(0-(-30))\\Q_{1}=18810 [J]\\Q_{2}=0.3*334*10^{3} \\Q_{2}=100200[J]\\Q_{3}=0.3*4180*(100-0)\\Q_{3}=125400[J]\\Q_{4}=0.3*2260*10^{3}\\Q_{4}=678000[J]\\Q_{5}=0.3*2010*(130-100)\\Q_{5}=18090[J]$

And the final step is the sum of all the five heat processes.

$Q_{t}=Q_{1}+Q_{2}+Q_{3}+Q_{4}+Q_{5}\\Q_{t}=18810+100200+125400+678000+18090\\Q_{t}=940500[J]$