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?

A) 1867.0 J
B) 2260.4 J
C) 7897.0 J
D) 42657.0 J

Q = 2260.44 j

Explanation:

Given data:

Mass of ice = 18 g

Initial temperature = -10 °C

Final temperature = 20°C

Heat absorbed by ice  = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

ΔT = 20°C - (-10°C)

ΔT =  30°C

Specific heat of water = 2.00 j/g.°C

Q = 18 g × 4.186 j/g.°C × 30°C

Q = 2260.44 j

58750.08068 is the answer i think