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Taking into account the definition of calorimetry, sensible heat and latent heat, the heat needed to melt 10.0 grams of ice at -10°C until it is water at 10°C is 3,969.5 J.
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).
In this case, 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.
- c is a specific heat substance.
- ΔT is the temperature variation.
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.
In this case, 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.
This case
First, you have to get the ice from -10°C to 0 °C, which is the melting point. Then you have to melt the ice into liquid water. Now, you have to get the water from 0°C to 10°C.
-10°C to 0 °C
In this case, you know:
- c= specific heat capacity of ice= 2.108 [tex]\frac{J}{gK}[/tex]
- m= 10 g
- ΔT= Tfinal - Tinitial= 0 °C - (-10 °C)= 10 °C= 10 K because being a temperature difference, the difference is the same in °C and K.
Replacing in the definition of sensible heat:
Q1= 2.108 [tex]\frac{J}{gK}[/tex]× 10 g× 10 K
Solving:
Q1= 210.8 J
Heat needed to melt ice
In this case, you have to melt the ice into liquid water. Being the specific heat of melting of ice is 334 J/g, the heat needed to melt 10 grams of ice is calculated as:
Q2= 10 grams× 334 J/g
Solving:
Q2= 3,340 J
0°C to 10 °C
In this case, you know:
- c= specific heat capacity of liquid water is 4.187 [tex]\frac{J}{gK}[/tex]
- m= 10 g
- ΔT= Tfinal - Tinitial= 10 °C - 0 °C= 10 °C= 10 K because being a temperature difference, the difference is the same in °C and K.
Replacing in the definition of sensible heat:
Q3= 4.187 [tex]\frac{J}{gK}[/tex]× 10 g× 10 K
Solving:
Q3= 418.7J
Total heat required
The total heat required is calculated as:
Total heat required= Q1 + Q2 + Q3
Total heat required= 210.8 J + 3,340 J + 418.7 J
Total heat required= 3,969.5 J
In summary, the heat needed to melt 10.0 grams of ice at -10°C until it is water at 10°C is 3,969.5 J.
Learn more about calorimetry:
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