The specific heat capacity of ice is 2.032 J/g degree C. If the initial temperat

Question

The specific heat capacity of ice is 2.032 J/g degree C. If the initial temperature of a 1.00 square meter patch of ice is -6 degree C, what is its final temperature after being in sunlight for 12 hours, assuming no phase changes and assuming that sunlight penetrates uniformly to a depth of 1.00 cm? what mass of carbon dioxide is present in 1.00 m^3 of day air at a temperature of 21 degree C and a pressure of 717 torr? Express your answer with the appropriate units. Calculate the mass percentage of oxygen in dry air. Express your answer with the appropriate units.

Explanation / Answer

Sunlight striking Earth's surface supplies 168 W per square meter. hence for 1square meter, the heat supplied= 168 W=168 J/s. for 12 hours, it is 168*12*60*60 Joules =7257600 joules

Volume of ice = 1m2*1/100m = 0.01 m3. mass of ice = density* volume = 0.01*1000 =10kg =10*1000 gm

hence mass*specific heat*temperature difference =7257600

10*1000*2.02*(t+6.0)= 7257600, t= 353 deg.c

2. Air contains 400ppm= 0.4% CO2 by volume in air

moles of air in 1m3=1000L , T=21+273=294, P= 717/760 atm =0.943 atm and R=0.0821 L.atm/mole.k

n= no of moles of air = PV/RT= 0.943*1000/(0.0821*294)= 39 moles

moles of CO2 in 39 moles =39*0.4/100 = 0.156 moles. mass of CO2= moles* molar mass= 0.156*44= 0.864 gm

3. Dir air contains 21% O2 and 79% N2 by moles

Basis : 1 mole of air. It contains 0.21 mole of O2, 0.79 moles of N2

Mass of O2= moles* molar mass =0.21*32= 6.72 gm, mass of N2=0.79*28= 22.12

mass of 1 mole of air =6.72+22.12= 28.84, mass % of air = 100* mass of O2/ total mass= 100*6.72/28.84=23.3%

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