Question

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?

Answer 1

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.

Answer 2

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