Respuesta :
Taking into account the definition of calorimetry, sensible heat and latent heat, the amount of heat required is 463.28 kJ.
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.
Energy required in this case
- -25°C to 0 °C
In firts place, the melting point of water (temperature at which it changes state from solid to liquid) is 0°C.
So, first of all you must increase the temperature from -25 ° 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(ice)= 2050 [tex]\frac{J}{kgC}[/tex]
- m= 845 g= 0.845 kg
- ΔT= Tfinal - Tinitial= 0 °C - (-25) °C= 25 °C
Replacing:
Q1= 2050 [tex]\frac{J}{kgC}[/tex] × 0.845 kg× 25 °C
Solving:
Q1= 43,306.25 = 43.3 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= 0.845 kg
- ΔHfus= 334 [tex]\frac{kJ}{kg}[/tex]
Replacing:
Q2= 0.845 kg× 334[tex]\frac{kJ}{kg}[/tex]
Solving:
Q2= 282.23 KJ
- 0 °C to 39 °C
Similar to sensible heat previously calculated, you know:
- c(liquid)= 4180[tex]\frac{J}{kgC}[/tex]
- m= 0.845 kg
- ΔT= Tfinal - Tinitial= 39 °C - 0 °C= 39 °C
Replacing:
Q3= 4180 [tex]\frac{J}{kgC}[/tex] × 0.845 kg× 39 °C
Solving:
Q3= 137,751.9 J= 137.75 kJ
- Total heat required
The total heat required is calculated as:
Total heat required= 43.3 kJ + 282.23 kJ + 137.75 kJ
Total heat required= 463.28 kJ
In summary, the amount of heat required is 463.28 kJ.
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The amount of energy required if the final temperature of the system is to reach 39 °C is 463288.15 J
How to determine the heat required change the temperature from –25 °C to 0 °C
- Mass (M) = 845 g = 845 / 1000 = 0.845 Kg
- Initial temperature (T₁) = –25 °C
- Final temperature (T₂) = 0 °
- Change in temperature (ΔT) = 0 – (–25) = 25 °C
- Specific heat capacity (C) = 2050 J/(kg·°C)
- Heat (Q₁) =?
Q = MCΔT
Q₁ = 0.845 × 2050 × 25
Q₁ = 43306.25 J
How to determine the heat required to melt the ice at 0 °
- Mass (m) = 0.845 Kg
- Latent heat of fusion (L) = 334 KJ/Kg = 334 × 1000 = 334000 J/Kg
- Heat (Q₂) =?
Q = mL
Q₂ = 0.845 × 334000
Q₂ = 282230 J
How to determine the heat required to change the temperature from 0 °C to 39 °C
- Mass (M) = 0.845 Kg
- Initial temperature (T₁) = 0 °C
- Final temperature (T₂) = 39 °
- Change in temperature (ΔT) = 39 – 0 = 39 °C
- Specific heat capacity (C) = 4180 J/(kg·°C)
- Heat (Q₃) =?
Q = MCΔT
Q₃ = 0.845 × 4180 × 39
Q₃ = 137751.9 J
How to determine the heat required to change the temperature from –25 °C to 39 °C
- Heat for –25 °C to 0°C (Q₁) = 43306.25 J
- Heat for melting (Q₂) = 282230 J
- Heat for 0 °C to 39 °C (Q₃) = 137751.9 J
- Heat for –25 °C to 39 °C (Qₜ) =?
Qₜ = Q₁ + Q₂ + Q₃
Qₜ = 43306.25 + 282230 + 137751.9
Qₜ = 463288.15 J
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