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?

The equation to be used to
calculate the heat absorbed when solid melts is the enthalpy of melting. It is
represented as H
= mCpT where H is the heat
released after phase change, m is the mass of the substance, Cp is the specific
heat capacity of the substance and T
is the change in temperature.

H
= mCpT

H
= (18g/1000g/kg)(4.184kJ/kg-K)(293-(-263))

H
= 41.87kJ

Answer Link

keniaojedax3 keniaojedax3 · Answer 2 · 2020-04-14T21:03:22-04:00

Answer:

7897.0 J

Explanation:

The correct answer is 7897.0 J. Break this problem down into three parts. First, how much energy is required to bring the ice from -10°C to 0°C = mCΔT = 18g x 2

J

g°C

x 10°C = 360 J. Add to this the amount of energy required to melt the ice (note that the temperature always stays constant during a phase change). The number of moles is 1, since the mass is 18g. mol x ΔHFus = 1mol x 6030

J

mol

= 6030J. Finally, add to this the amount of energy required to heat the (now liquid) water from 0 to 20°C = mCΔT = 18g x 4.186

J

g°C

x 20°C = 1507 J. To get the final answer, simply add the three steps together = 360J + 6030J + 1507J = 7897J.