A) Find the heat needed to change 1.6 kg of ice to water at 0 degree Celsius. B) Calculate the...


A) Find the heat needed to change {eq}1.6 \, \mathrm{kg} {/eq} of ice to water at {eq}0 ^{\circ} C {/eq}.

B) Calculate the total energy needed to evaporate completely {eq}1 \, \mathrm{kg} {/eq} of ice at {eq}-10 ^{\circ} C {/eq}.

C) A heater marked {eq}60 \, \mathrm{W} {/eq} evaporates {eq}6 \times 10^{-3} \, \mathrm{kg} {/eq} of boiling water in {eq}60 {/eq} seconds. What is the specific latent heat of vaporization of water?

Heating Curve/Coolilng Curve

Heating curve and cooling curve plots the temperature on the y-axis and time (or amount of heat transferred) on the x-axis. The sloped regions are when one phase exits the energy transfer is reflected in the temperature change. The plateau regions are when two phases are involved and energy is used to change one phase to another without temperature change.

Answer and Explanation:

3. 530 kJ

4. 3200 kJ

5. {eq}6 \times 10^5\; J/kg {/eq}

3. A phase change is taking place without temperature change and so we need to use the enthalpy of fusion for {eq}H_2O {/eq} ({eq}\Delta H_{fus, \;H_2O} = 6.02\; kJ/mole {/eq})as shown below

$$1.6\;kg \;H_2O \times \dfrac {1000\;g\;H_2O}{1\;kg\;H_2O} \times \dfrac {1\;mole\;H_2O}{18.02\;g\;H_2O} \times \dfrac {6.02\; kJ\;H_2O}{1\;mole\;H_2O} = \boxed {530\;kJ} $$

4. Both temperature changes (using specific heats) and phase changes (using enthalpy of phase changes) are involved.

  • Heating Ice from -10 to 0 {eq}^oC {/eq}

$$(1000\;g)(2.09\; \dfrac {J}{g \cdot ^oC})(0- -10\;^oC) =20900\;J = 20.9\;kJ $$

  • Melting ice at 0 {eq}^oC {/eq}

$$(1000\;g)\times \dfrac {1\;mole\;H_2O}{18.02\;g\;H_2O} \times \dfrac {6.02\; kJ\;H_2O}{1\;mole\;H_2O} = 334\;kJ $$

  • Heating water from 0 to 100 {eq}^oC {/eq}

$$(1000\;g)(4.18\; \dfrac {J}{g \cdot ^oC})(100-0\;^oC) =418000\;J = 418\;kJ $$

  • Evaporate water at 100 {eq}^oC {/eq}

$$(1000\;g)\times \dfrac {1\;mole\;H_2O}{18.02\;g\;H_2O} \times \dfrac {44.0\; kJ\;H_2O}{1\;mole\;H_2O} = 2442\;kJ $$

The total energy involved would be the sum of all the steps

$$20.9\;kJ + 334\;kJ + 418\;kJ + 2442\;kJ = \boxed {3200\;kJ} $$

5. Specific latent heat has unit of J/kg

$$(60\;W)(60\;s) = 3600\;J\\ \dfrac {3600\;J}{6 \times 10^{-3}\;kg} = \boxed {6 \times 10^5\; J/kg} $$

Learn more about this topic:

Phase Changes and Heating Curves

from Chemistry 101: General Chemistry

Chapter 6 / Lesson 3

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