Taking into account the definition of calorimetry, sensible heat and latent heat, the amount of heat required is 37.88 kJ.
<h3>Calorimetry</h3>
Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
<h3>Sensible heat</h3>
Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).
<h3>Latent heat</h3>
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.
- <u><em>25.60 °C to 0 °C</em></u>
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
- m= 185.5 g
- ΔT= Tfinal - Tinitial= 0 °C - 25.60 °C= - 25.6 °C
Replacing:
Q1= 4.184 × 185.5 g× (- 25.6 °C)
Solving:
<u><em>Q1= -19,868.98 J</em></u>
- <u><em>Change of state</em></u>
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× 
= 10.30 moles, where 18 
is the molar mass of water, that is, the amount of mass that a substance contains in one mole.
ΔHfus= 6.01 
Replacing:
Q2= 10.30 moles×6.01
Solving:
<u><em>Q2=61.903 kJ= 61,903 J</em></u>
- <u><em>0 °C to -10.70 °C</em></u>
Similar to sensible heat previously calculated, you know:
- c = Heat Capacity of Solid = 2.092
- m= 185.5 g
- ΔT= Tfinal - Tinitial= -10.70 °C - 0 °C= -10.70 °C
Replacing:
Q3= 2.092 × 185.5 g× (-10.70) °C
Solving:
<u><em>Q3= -4,152.3062 J</em></u>
<h3>Total heat required</h3>
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
<u><em>Total heat required= 37,881.7138 J= 37.8817138 kJ= 37.88 kJ</em></u>
In summary, the amount of heat required is 37.88 kJ.
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