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  • Given: - The heat of fusion for a substance is <img alt="250 \mathrm{~J} / \mathrm{g}." src="http

Given: - The heat of fusion for a substance \mathrm{Y}is 250 \mathrm{~J} / \mathrm{g}.

- The heat of vaporization for substance Y is 1200 \mathrm{~J} / \mathrm{g}.

- The molar mass of substance \mathrm{Y} is \mathrm{80 \mathrm{~g} / \mathrm{mol}.}

- Initial temperature of substance \mathrm{Y} is \mathrm{-20^{\circ} \mathrm{C}.}

- Final temperature of the vaporized substance \mathrm{Y} is 150^{\circ} \mathrm{C}.

- The specific heat capacity of substance Y (liquid) is 4.0 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}.

- The specific heat capacity of vaporized substance \mathrm{Y} is 2.5 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}.

Calculate the enthalpy change for the following process: Heating 40 \mathrm{~g} of substance \mathrm{Y} at -20^{\circ} \mathrm{C} to vaporized substance \mathrm{Y} at \mathrm{150^{\circ} \mathrm{C}}.

Option: 1

37.2 kJ


Option: 2

-37.2 kJ


Option: 3

-500 kJ


Option: 4

-250 kJ


Answers (1)

best_answer

Step 1: Calculate the heat required to raise the temperature of substance Y from \mathrm{-20^{\circ} \mathrm{C} \, \, to \, \, 0^{\circ} \mathrm{C}.}

                                                      \mathrm{ q_1=m \times C_{\mathrm{Y}, \text { liquid }} \times \Delta T }
Where: -m=40 \mathrm{~g} (mass of substance \mathrm{Y} ) \mathrm{-C_{\mathrm{Y}}, liquid =4.0 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}} (specific heat capacity of substance \mathrm{Y})-\Delta T=0-(-20)^{\circ} \mathrm{C}=20^{\circ} \mathrm{C}

                                                       \mathrm{ q_1=40 \mathrm{~g} \times 4.0 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C} \times 20^{\circ} \mathrm{C}=3200 \mathrm{~J} }
Step 2: Calculate the heat required to melt substance \mathrm{Y}at 0^{\circ} \mathrm{C} to substance $\mathrm{Y}$ at 0^{\circ} \mathrm{C}.

                                                        \mathrm{ q_2=m \times \Delta H_{\text {fusion }} }
Where: -m=40 \mathrm{~g} (mass of substance \mathrm{Y} ) -\Delta H_{\text {fusion }}=250 \mathrm{~J} / \mathrm{g} (heat of fusion for substance \mathrm{Y} )

                                                        \mathrm{ q_2=40 \mathrm{~g} \times 250 \mathrm{~J} / \mathrm{g}=10000 \mathrm{~J} }
Step 3: Calculate the heat required to raise the temperature of substance \mathrm{Y} from \mathrm{0^{\circ} \mathrm{C}} to \mathrm{150^{\circ} \mathrm{C}.}

                                                          \mathrm{ q_3=m \times C_{\mathrm{Y}, \text { liquid }} \times \Delta T }

Where: -m=40 \mathrm{~g} (mass of substance \mathrm{Y}) \mathrm{-C_{\mathrm{Y}}}, liquid \mathrm{=4.0 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}} (specific heat capacity of substance \mathrm{Y})-\Delta T=150^{\circ} \mathrm{C}-0^{\circ} \mathrm{C}=150^{\circ} \mathrm{C}

                                                       \mathrm{ q_3=40 \mathrm{~g} \times 4.0 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C} \times 150^{\circ} \mathrm{C}=24000 \mathrm{~J} }

The total heat for the process is q_{\text {total }}=q_1+q_2+q_3=3200 \mathrm{~J}+10000 \mathrm{~J}+ 24000 J=37.2 k J

Posted by

avinash.dongre

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