Part A
How much heat is required to convert 4.88 g of ice at-14.0°C to water at 23.0°C? (The heat capacity of ice is 2.09 J/(g- "C). AHp (H₂O) = 40.7 kJ/mol, and AH (H₂O) = 6.02 kJ/mol.)
Express your answer with the appropriate units.
?

Respuesta :

Taking into account the definition of calorimetry, sensible heat and latent heat, the amount of heat required is 2.243974 kJ or 2,243.974 J.

Definition of calorimetry, sensible heat and latent heat

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

Latent heat is defined as the energy required by a quantity of substance to change state.

When this change consists of changing from a solid to a liquid phase, it is called heat of fusion and when the change occurs from a liquid to a gaseous state, it is called heat of vaporization.

  • -14 °C to 0°C

In firts place, you know that the melting point is 0°C. So, first of all you must increase the temperature from -14 ° C (in solid state) to 0 ° C, in order to supply heat without changing state (sensible heat).

The amount of heat a body receives or transmits is determined by:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

  • c(solid)= heat capacity of ice= 2.09 [tex]\frac{J}{gC}[/tex]
  • m= 4.88 g
  • ΔT= Tfinal - Tinitial= 0 °C - (-14) °C= 14 °C

Replacing:

Q1= 2.09 [tex]\frac{J}{gC}[/tex] × 4.88 g× 14 °C

Solving:

Q1=142.7888 J= 0.1427888 kJ

  • Change of state

The heat Q that is necessary to provide for a mass m of a certain substance to change phase is equal to

Q = m×L

where L is called the latent heat of the substance and depends on the type of phase change.

In this case, you know:

  • n= 4.88 g×[tex]\frac{1 mole}{18 g}[/tex] = 0.2711 moles, where 18 [tex]\frac{g}{mole}[/tex] is the molar mass of H₂O, that is, the amount of mass that a substance contains in one mole.
  • ΔHfus= 6.02 [tex]\frac{kJ}{mol}[/tex]

Replacing:

Q2= 0.2711 moles×6.02 [tex]\frac{kJ}{mol}[/tex]

Solving:

Q2= 1.632022 kJ

  • 0 °C to 23 °C

Similar to sensible heat previously calculated, you know:

  • c(liquid)= specific heat of water= 4.18[tex]\frac{J}{gC}[/tex]
  • m= 4.88 g
  • ΔT= Tfinal - Tinitial= 23 °C - 0 °C= 23 °C

Replacing:

Q3= 4.18[tex]\frac{J}{gC}[/tex] × 4.88 g× 23 °C

Solving:

Q3= 469.1632 J= 0.4691632 kJ

  • Total heat required

The total heat required is calculated as:  

Total heat required= 0.1427888 kJ + 1.632022 kJ + 0.4691632 kJ

Total heat required= 2.243974 kJ= 2,243.974 J

In summary, the amount of heat required is 2.243974 kJ or 2,243.974 J.

Learn more about calorimetry:

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