Respuesta :
Answer:
The specific heat of alloy [tex]C_{alloy} = 1.007 \frac{KJ}{Kg K}[/tex]
Explanation:
Mass of the alloy = 30.5 gm = 0.0305 kg
Initial temperature = 93 °c = 366 K
Mass of water = 50 gm = 0.05 kg
Initial temperature = 22 °c = 295 K
Final temperature of the mixture = 31.1 °c = 304.1 K
From the energy conservation principal the heat lost by the alloy is equal to heat gain by the water.
Heat lost by alloy
[tex]Q_{alloy} = m C (T_{f}- T_{i} )[/tex]
[tex]Q_{alloy} = (0.0305) C_{alloy} (366-304.1)[/tex]
[tex]Q_{alloy} = (1.88795) C_{alloy}[/tex] ------- (1)
Heat gain by water
[tex]Q_{w} = (0.05) (4.18) (304.1 - 295)[/tex]
[tex]Q_{w} = 1.9019 \frac{KJ}{kg K}[/tex] ------- (2)
Equation (1) = Equation (2)
[tex](1.88795) C_{alloy} = 1.9019[/tex]
[tex]C_{alloy} = 1.007 \frac{KJ}{Kg K}[/tex]
This is the specific heat of alloy.
Taking into account the definition of calorimetry, the specific heat capacity of the alloy is 1.063 [tex]\frac{kJ }{kgK}[/tex]
In first place, calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.
In this way, between heat and temperature there is a direct proportional relationship.
The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat by the mass of the body.
So, the equation that allows to calculate heat exchanges is:
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:
for alloy:
- Calloy= ?
- m= 30.5 g= 0.0305 kg (being 1000 g= 1 kg)
- ΔT=Tfinal - Tinitial= 93 C - 31.1 C= 61.9 C= 61.9 K because it is a temperature difference, it has the same value in both units
for water:
- Cwater= 4.18 [tex]\frac{kJ}{kgK}[/tex]
- m= 50 g= 0.050 kg (being 1000 g= 1 kg)
- ΔT=Tfinal - Tinitial= 31.1 C - 22 C= 9.2 C= 9.2 K
Replacing in the expression to calculate heat exchanges:
for alloy:
- Qalloy= Calloy× 0.0305 kg× 61.9 K
for water:
- Qwater= 4.18 [tex]\frac{kJ}{kgK}[/tex]× 0.050 kg× 9.2 K
On the other hand, the heat of the calorimeter can be expressed as:
Qcalorimeter= Ccalorimeter×ΔT
Being:
- Ccalorimeter= 9.2 [tex]\frac{J}{K}[/tex]= 0.0092 [tex]\frac{kJ}{K}[/tex]
- ΔT=Tfinal - Tinitial= 31.1 C - 22 C= 9.2 C= 9.2 K
and replacing you get:
Qcalorimeter= 0.0092 [tex]\frac{kJ}{K}[/tex]× 9.2 K
It should be taken into account that a system at different temperatures evolves spontaneously towards a state of equilibrium in which all bodies have the same temperature. Then, mixing two quantities of liquids at different temperatures generates an energy transfer in the form of heat from the hottest to the coldest. Said energy transit is held until temperatures equalize, when it is said to have reached thermal equilibrium.
So the heat released by the sample is absorbed by the calorimeter and the water.
Qalloy= Qcalorimeter + Qwater
Replacing the corresponding expressions and solving:
Calloy× 0.0305 kg× 61.9 K= 0.0092 [tex]\frac{kJ}{K}[/tex]× 9.2 K + 4.18 [tex]\frac{kJ}{kgK}[/tex]× 0.050 kg× 9.2 K
Calloy× 1.88795 kg× K= 0.08464 kJ + 1.9228 kJ
Calloy× 1.88795 kg× K= 2.00744 kJ
[tex]Calloy=\frac{2.00744 kJ }{1.88795 kgK}[/tex]
Calloy= 1.063 [tex]\frac{kJ }{kgK}[/tex]
Finally, the specific heat capacity of the alloy is 1.063
Learn more:
- brainly.com/question/11586486?referrer=searchResults
- brainly.com/question/24724338?referrer=searchResults
