Question

A) How much heat energy, in kilojules, is required to covert 33.0g of ice at -18.0 degrees Celsius to water at 25.0 degrees Celsius? Express answer to 3 dig figs with units.

B) How long would it take for 1.50 mol of water at 100.0 degrees Celsius to be converted completely into steam if heat were added at a constant rate of 24.0J/s? Express answer to 3 dig figs with units.

Answer 1

A)

Specific Heat of Ice = C_ice = 2.1 J/g/°C. Latent heat of melting ice = = 334 J/g. Specific Heat of water = C_water = 4.184J/g/°C.

These are the constants we have to use for this problem

First step:

Heat required to convert ice from -18 to 0 ^oC =

Initial temperature,T₁ = -18 ^oC

Final temperature T₂ = 0 ^oC

Q₁ = m*C_ice*(T₂-T₁)

Q₁ = 33.0 * 2.1* (0 - (-18))

Q₁ = 33.0 * 2.1* (18)

Q₁ = 1247.4 J

Second step:

Heat required to melt ice at 0 ^oC = Q₂ = m x =33.0 g x 334 J/g = 11,022 J

Step 3:

Heat required to heat the water from 0 to 25.0 deg C =

Q₃ = m*C_water*(T₂-T₁)

Q₃ = 33 * 4.184 * (20)

Q₃ = 2761.4 J

Total heat required = Q₁ + Q₂ + Q₃ = 1247.4 J +11,022 J + 2761.4 J = 15030.8 J

Total heat required = 15.0 kJ

B)

Latent heat of vaporization of water at 100 ^oC = 2258 kJ/kg

Moles of water = 1.50 mol

Mass of water =

Mass of water = 27 g

Enthanlpy of Vaporization =

Heat is supplied at =24.0 J/s

Time required for complete vaporization of water =

Time take for 1.50 mol of water at 100.0 degrees Celsius to be converted completely into steam = 42.0 min