# 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.

- 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.
- 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.)
- 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?

### 1 Answer | Add Yours

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**

**Sources:**