Specific heat of water = 4.186 j g°c specific heat of ice = 2.00 j g°c molar heat of fusion = 6030 j mol molar heat of vaporization = 40790 j mol you take an ice cube (mass = 18g) from the freezer (t = -10°c) and place it on the table. later that day, you notice a puddle of water on the table that has reached ambient room temperature (20°c). how much heat must have been absorbed to make this happen?

8184.96 J

The ice would reach a temperature 0° C. It would melt to form water and again gain heat to reach ambient room temperature of 20° C.

Therefore, the heat would be absorbed for the three processes,

The rise of temperature from -18°C to 0°C, Q₁= m c ΔT

where, c is specific heat of ice

Q₁= m c ΔT = 18 g × 2.00 J/g° C × 18° C = 648 J

Melting of ice at 0°C, Q₂ = m L = 1 mol ×  6030  J/mol = 6030 J

where L is the latent heat of fusion

The rise of temperature of water from 0°C to 20°C, Q₃= m c' ΔT'

Where, c' is the specific heat of water.

Q₃= 18 g ×  4.186 J × 20° C = 1506.96 J

Net heat absorbed:

Q = Q₁ + Q₂ + Q₃ =  648 J+6030 J+1506.96 J = 8184.96 J

Hence, 8184.96 J heat must be absorbed to make this happen.

the arctic is a region of the planet, north of the arctic circle, and includes the arctic ocean, greenland, baffin island, other smaller northern islands, and the far northern parts of europe, russia (siberia), alaska and canada.

i think it would be 5