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.