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Taking into account the definition of calorimetry, sensible heat and latent heat, the amount of heat required is 37.88 kJ.
Calorimetry
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
Sensible heat
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
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.
- 25.60 °C to 0 °C
First of all, you should know that the freezing point of water is 0°C. That is, at 0°C, water freezes and turns into ice.
So, you must lower the temperature from 25.60°C (in liquid 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= Heat Capacity of Liquid= 4.184 [tex]\frac{J}{gC}[/tex]
- m= 185.5 g
- ΔT= Tfinal - Tinitial= 0 °C - 25.60 °C= - 25.6 °C
Replacing:
Q1= 4.184 [tex]\frac{J}{gC}[/tex]× 185.5 g× (- 25.6 °C)
Solving:
Q1= -19,868.98 J
- 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= 185.5 grams× [tex]\frac{1mol}{18 grams}[/tex]= 10.30 moles, where 18 [tex]\frac{g}{mol}[/tex] is the molar mass of water, that is, the amount of mass that a substance contains in one mole.
ΔHfus= 6.01 [tex]\frac{kJ}{mol}[/tex]
Replacing:
Q2= 10.30 moles×6.01 [tex]\frac{kJ}{mol}[/tex]
Solving:
Q2=61.903 kJ= 61,903 J
- 0 °C to -10.70 °C
Similar to sensible heat previously calculated, you know:
- c = Heat Capacity of Solid = 2.092 [tex]\frac{J}{gC}[/tex]
- m= 185.5 g
- ΔT= Tfinal - Tinitial= -10.70 °C - 0 °C= -10.70 °C
Replacing:
Q3= 2.092 [tex]\frac{J}{gC}[/tex] × 185.5 g× (-10.70) °C
Solving:
Q3= -4,152.3062 J
Total heat required
The total heat required is calculated as:
Total heat required= Q1 + Q2 +Q3
Total heat required=-19,868.98 J + 61,903 J -4,152.3062 J
Total heat required= 37,881.7138 J= 37.8817138 kJ= 37.88 kJ
In summary, the amount of heat required is 37.88 kJ.
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