Question

(a). The world ocean absorbed 193.0 x 10^21 Joules of heat between 1972 and 2008. Estimate what must have been the average rate of global ocean temperature increase over this period due to that heat input in degrees C/ year.

(b).Related to this increase in ocean temperature, global-mean sea level rose by an estimated 33 +/- 12mm from 1972 - 2008 due to thermal expansion. Now imagine that, rather than being applied to the oceans, this 193.0 x 10^21 Joules of heat had instead been applied to the land ice on earth. What mass of land ice would have melted and how much would sea levels have risen in this case? (You can assume that all of the land ice begins at a temperature of 0 degrees C).

Useful values:

Mass of Ocean = 1.4 x 10^21 kg Mass of ice on Earth = 2.2 x 10^19 kg Area of ocean = 3.61 x 10^14 m^2

Salinity of Seawater = 35% Density of Seawater = 1035 kg/m^3 Density of Fresh Water = 1000 kg/m^3

Specific Heat of Seawater = 3.994 J/g/deg C Latent heat of fusion of water/ice = 334 J/g/deg C

Answer 1

heat absorbed by world ocean = 193*10²¹ joules

Specifc heat of sea water= 3.994 j/g.deg.c

Mass of ocean = 1.4*1021 kg

Temperature rise= heat absorbed/ ( specific heat* mass of ocean)= 193*10²¹/ (3.994*1.4*10²¹) =34.51 deg.c

b) Heat absorbed by ocean =193*10²¹ joules, Latent heat of fusion of ice= 334 J/g

Mass of ice melted= 193*10²¹/ 334=5.8*10²⁰ gms= 5.8*10¹⁷ kg

Volume of land ice = mass/density= 5.8*10¹⁷/ 1000=5.8*10¹⁴ m3

Area of sea =3.61*10¹⁴ m2

level rise in sea= 5.8/3.61=1.61 m