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Stells [14]
3 years ago
10

En un experimento de calorimetría, 0.50 kg de un metal a 100°C se añaden a 0.50 kg de agua a 20°C en un vaso de calorímetro de a

luminio, cuya masa es de 0.250 kg. A) Si un poco de agua salpica y sale del vaso al agregar el metal, el calor específico medido será 1) mayor, 2) igual o 3) menor que el valor calculado para el caso en que no se salpique agua. ¿Por qué? B) Si la temperatura final de la mezcla es de 25°C, y no se salpica agua, ¿qué calor específico tendrá el metal?
Physics
1 answer:
Maru [420]3 years ago
7 0

Answer:

c=0.14J/gC

Explanation:

A.

2) The specific heat will be the same because it is a property of the substance and does not depend on the medium.

B.

We can use the expression for heat transmission

Q=mc(T_2-T_1)

In this case the heat given by the metal (which is at a higher temperature) is equal to that gained by the water, that is to say

Q_1=-Q_2

for water we have to

c = 4.18J / g ° C

replacing we have

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

I hope this is useful for you

A.

2) El calor específico será igual porque es una propiedad de la sustancia y no depende del medio.

B.

Podemos usar la expresión para la transmisión de calor

Q=mc(T_2-T_1)

En este caso el calor cedido por el metal (que está a mayor temperatura) es igual al ganado por el agua, es decir

Q_1=-Q_2

para el agua tenemos que

c=4.18J/g°C

reemplazando tenemos

c_{metal}*(500g)(100\°C-25\°C)=-(250g)(4.18\frac{J}{g\°C})(20\°C-25\°C)\\c_{metal}=0.14\frac{J}{g\°C}

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Pls see attached file

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Help !!!! Estimate the number of breaths taken by a person during 44 years?
Lisa [10]
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x
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6 0
2 years ago
If 10 waves pass a point each second and their wavelength is 30m, what is their speed?
svet-max [94.6K]

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3 years ago
To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V
aleksley [76]

Answer:

The maximum energy stored in the combination is 0.0466Joules

Explanation:

The question is incomplete. Here is the complete question.

Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.

Energy stored in a capacitor is expressed as E = 1/2CtV² where

Ct is the total effective capacitance

V is the supply voltage

Since the capacitors are connected in series.

1/Ct = 1/C1+1/C2+1/C3

Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF

1/Ct = 1/11.7 + 1/21.0 + 1/28.8

1/Ct = 0.0855+0.0476+0.0347

1/Ct = 0.1678

Ct = 1/0.1678

Ct = 5.96μF

Ct = 5.96×10^-6F

Since V = 125V

E = 1/2(5.96×10^-6)(125)²

E = 0.0466Joules

8 0
2 years ago
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