A winter outdoors man awakens one January morning at his campsite in the Adirond
ID: 2075409 • Letter: A
Question
A winter outdoors man awakens one January morning at his campsite in the Adirondack Mountains of upstate New York. He is at approximately 400m of elevation, and the outside temperature is ten degrees below zero Fahrenheit. Wanting a hot cup of coffee, he places 10 lbm of solid ice into pot, and places the pot over a fire to heat.
1) Calculate the sensible heat needed to bring the ice to melting.
2) Calculate the total heat of fusion needed to melt the ice.
3) Calculate the sensible heat needed to bring the water to the desired coffee temperature.
Once the water reaches 98 C, he removes 0.25 lbm from the pot to make a cup of coffee.
4) Once he makes his coffee, the water continues to heat until boiling. The camper gets distracted when he sees a large buck running in the woods, and forgets to take the pot off the fire. When he returns, he finds the pot empty. Assuming 0.25 lbm of water left as vapor before boiling began, how much energy did it take to boil away the remaining water?
5) How much TOTAL energy was transferred to the water?
Explanation / Answer
State conditions:
Temperature= -10 F= -23.33 C
altitude=400m above sea level
therefore pressure= 95903.84 Pa
mass of ice =10 lbm=4.5359 kg
Find the melting temperature, boiling temperature and specific heats at given state conditions
At lower atmospheric pressures the melting and boiling point decreases from standard 0 K and 273K
but the variation is slight hence in this calculation is been neglected. If required follow Clausius Clapeyron relation
1) sensible heat
Specific heat of ice= 2.108 KJ/kg K
Q sensible =m C dT
=4.5359*2.108*23.33
=223.0739KJ
2) total heat of fusion
Q fusion= Q sensible +mL
=233.0739+(4.5359*334*)
=1748.0645 KJ
3) Tf= 98 C
total heat= Q fusion+M*Cwater dT
3606.728KJ
4) water left =10-.25=9.75 lbm
=4.4225 kg
0.25 lbm evaporated before boiling
therefore, water left in pot=4.309kg
energy required to boil remaining water
initial temperature 98C final 100C
Q= mCdT + mL
=m(CdT+L)
=4.309(4.186(2)+2264.76)
9794.925KJ
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