How much heat is required to change a 36.8 g ice cube from ice at -14.9°C to wat
ID: 1476719 • Letter: H
Question
How much heat is required to change a 36.8 g ice cube from ice at -14.9°C to water at 35°C? (if necessary, use cice=2090 J/kg°C and csteam= 2010 J/kg°C)How much heat is required to change a 36.8 g ice cube from ice at -14.9°C to steam at 120°C?
How much heat is required to change a 36.8 g ice cube from ice at -14.9°C to water at 35°C? (if necessary, use cice=2090 J/kg°C and csteam= 2010 J/kg°C)
How much heat is required to change a 36.8 g ice cube from ice at -14.9°C to steam at 120°C?
How much heat is required to change a 36.8 g ice cube from ice at -14.9°C to water at 35°C? (if necessary, use cice=2090 J/kg°C and csteam= 2010 J/kg°C)
How much heat is required to change a 36.8 g ice cube from ice at -14.9°C to steam at 120°C?
Explanation / Answer
36.8 g x (1 kg / 1000 g) = 0.0368 kg
First of all
1) The ice warms from -14.9ºC to its melting point, 0ºC.
Since this is a temperature change, we can use the formula q = mcT
q = (0.0368 kg)(2090 J/kgºC)(14.9ºC) = 1145 J
2) The ice melts while the temperature holds steady at 0ºC.
At this phase change, we can use the formula q = mL. we use the heat of fusion for L, which is 3.34x10^5 J/kg.
q = (0.0368 kg)(3.34x10^5 J/kg) = 12291 J
3) The meltwater warms from 0ºC to 35ºC.
Another temperature change, but this time it's water (rather than ice), so you'll need to use the specific heat of water.
q = (0.0368 kg)(4186 J/kgºC)(35ºC) = 5391 J
Add the three heats together and you get
q = 1145 J+12291 J+5391 J
q=18827 J
b)
1) The ice warms from -14.9ºC to its melting point, 0ºC.
Since this is a temperature change, we can use the formula q = mcT
q = (0.0368 kg)(2090 J/kgºC)(14.9ºC) = 1145 J
2) The ice melts while the temperature holds steady at 0ºC.
At this phase change, we can use the formula q = mL. we use the heat of fusion for L, which is 3.34x10^5 J/kg.
q = (0.0368 kg)(3.34x10^5 J/kg) = 12291 J
The meltwater warms from 0ºC to 100ºC.
3) Another temperature change, but this time it's water (rather than ice), so you'll need to use the specific heat of water instead.
q = (0.0368 kg)(4186 J/kgºC)(100ºC) =15404 J
4) The water boils while the temperature holds steady at 100ºC.
Another phase change. Use the heat of vaporization and the formula q = mL
q = (0.0368 kg)(2.26x10^6 J/kg) = 83168 J
5) The steam warms from 100ºC to 120ºC.
last temperature change. The specific heat of steam is about 2010 J/kgºC
q = (0.0368 kg)(2010 J/kgºC)(20ºC) = 1479 J
Now add the all temperature then we get the total energy
q=1145 J+12291 J+15404 J+83168 J+1479 J =113487J
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