# Calculate the mass of water held in both polar ice caps in grams. Assume that ice has a volume of 0.902 g/cm3.Consider the two substantial polar ice caps Greenland and Antarctica. It is estimated...

Calculate the mass of water held in both polar ice caps in grams. Assume that ice has a volume of 0.902 g/cm3.

Consider the two substantial polar ice caps Greenland and Antarctica. It is estimated that the volume of ice is 2.6 x 10^6 km3 for Greenland and 3.0 x 10^7 km^3 for Antarctica.

1. Calculate the mass of water held in both polar ice caps in grams. Assume that ice has a volume of 0.902 g/cm3. (There are 1x10^15 cm^3 in a km^3.) Show your work including any necessary formulas, all conversions, and all units.
2. Calculate the amount of heat in GJ required to melt all the ice in the polar ice caps at 0.0 °C. Assume that all the ice is held at –10 °C, (The specific heat of ice is 2.09 J/g x °C; the latent heat of fusion is 334 kJ/ kg; 1 Gigajoule = 10^9 joules.)
3. It is estimated that every 4x10^5 km3 of ice that melts will result in a sea level rise of 1.0 meter. Suppose that the polar ice caps absorb enough heat for 30.0 % of the ice caps to melt. How much will the sea level rise?

jeew-m | College Teacher | (Level 1) Educator Emeritus

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Density = mass/volume

Mass    = Density*Volume

1)

Mass of ice in Greenland

= `2.6*10^6*1*10^15*0.902 g`

= `2.345*10^21 g`

Mass of ice in Antarctica

= `3.0*10^7*1*10^15*0.902 g`

= `2.706*10^22 g`

So total ice mass in ice caps

= `2.345*10^21 +2.706*10^22`

= `2.94*10^22 g`

2)

When melting ice it will absorb the specific heat to come -10C to 0C. After that the ice will use its latent heat of fusion to melt to 0C water.

`Q = m*C*theta`

Q = heat absorbed in coming -10C ice to 0C ice

C = specific heat of ice

m = mass of ice

`theta` = temperature change

Heat required to come -10 to 0

= `2.94*10^22*2.09*(0-(-10))`

= `6.145*10^23 J`

= `6.145*10^23*10^-9 GJ`

= `6.145*10^14 GJ`

Q= m*l

Q = Heat absorbed in latent heat

m = mass of the ice

l = latent heat of fusion

Heat required to come 0 ice to 0 water

= `2.94*10^22*10^-3*334`

= `9.812*10^21 KJ`

= `9.812*10^21*10^-6`

= `9.812*10^15 GJ`

Heat required for the process

= `9.812*10^15+6.145*10^14 GJ`

=`1.043*10^16 GJ `

3)

Volume of ice in polar caps

= `2.6*10^6+3*10^7 Km^3`

= `3.26*10^7 km^3`

Melted ice volume

= 30% of total volume

= `0.3*3.26*10^7 km^3`

= `9.78*10^6 km^3`

sea level rise for melting of 4*10^5 km^3 ice = 1m

sea level rise for melting of 4*10^5 km^3 ice `

= 1/(4*10^5)*(9.78*10^6)`

= 24.45 m

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