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EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO
Msc. Widmar Aguilar
MAYO 2023
TEMPERATURA Y CALOR
EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO
Msc. Widmar Aguilar
MAYO 2023
EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO
Msc. Widmar Aguilar
MAYO 2023
EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO
Msc. Widmar Aguilar
MAYO 2023
Se tiene la relación:
.
=
.
= ; x-32 = 65.7
X = 32+65.7
X = 97.7 o
F
( )
=
= ; 32- x = 9
X = 32- 9
EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO
Msc. Widmar Aguilar
MAYO 2023
X = 23 o
F
b)
( )
=
= ; 32- x =54
X = 32- 54
X = -22 o
F
=
= ; x =66.67
X = 66.67 o
C
EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO
Msc. Widmar Aguilar
MAYO 2023
=
= ; x = - 3.89
X = -3.89 o
C
=
- = ; x = - 17.78
X = - 17.78 o
C
− 0
100
=
8 − 32
212 − 32
EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO
Msc. Widmar Aguilar
MAYO 2023
= ; x = - 13.33
X = -1 3.33 o
C
0 −
100
=
32 − (−50)
212 − 32
- = ; x = - 45.55
X = -45.55 o
C
0 −
100
=
32 −
212 − 32
- = ; -180 x = 3200 -100x
EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO
Msc. Widmar Aguilar
MAYO 2023
-80 X = 3200
X= -40
X = -40 o
C = - 40 o
F
50 − 0
100
=
− 273
373 − 273
= ; 50 = x-273
X = 323 o
K
Se sabe que: 0 o
C = 273 o
K
50 − 0
100
=
− 273
373 − 273
EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO
Msc. Widmar Aguilar
MAYO 2023
( )
= ; =
273-X = 35
X = 238 o
K
72 − 32
212 − 32
=
− 273
373 − 273
= ; x-273 = 22.22
X = 295.22 o
K
32 − 0
212 − 32
=
273 −
373 − 273
= ; 273 -x = 17.77
EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO
Msc. Widmar Aguilar
MAYO 2023
X = 255.22 o
K
32 − (−215)
212 − 32
=
273 −
373 − 273
= ; 273 -x = 137.22
X = 135.77 o
K
0 − (−273)
100 − 0
=
32 −
212 − 32
= ; 32-x = 491.4
X = - 459.4 o
F
EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO
Msc. Widmar Aguilar
MAYO 2023
CALOR:
Q = cm∆
C = 1 cal/g 0
C
Q = 1 cal/g 0
C * 350 g* (85-35) 0
C
Q = 350 * 50 cal
Q = 17500 cal
De: Q= c m ∆
0.95 cal = c (1g)(1) o
C
c = 0.95
!"#
Q= c m ∆ ; c = 0.032 cal /g*o
C
500 cal = 0.032 * 50 * (t-50)
t-50 = 312.5
t = 362.5 o
C
EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO
Msc. Widmar Aguilar
MAYO 2023
d = m/V
Q= c m ∆
Q = c d V ∆
Qhierro = 0.11 *7.87*V*10
Qplatino = 0.032 *21.45*V*10
$%&'(()
$*+,-&.)
=
0.11 ∗7.87∗V∗10
0.032 ∗21.45∗V∗10
$%&'(()
$*+,-&.)
=
0.11 ∗7.87
0.032 ∗21.45
= 1.26
Qhierro = 1.26 Qplatino
195 o
F = 90.55 o
C
284 o
K = 10.85 o
C
Qperdido = Q ganado
M1 c1 (∆ ) = M2 c2 (∆ ) : - Qperdido = Q gana
Qperdido = 200 g* 1 ca/g o
C *(t- 90.55)
200 g* 1 ca/g o
C *(90.55-t) = 120 g* 1 ca/g o
C *(t-10.85)
200*( 90.55-t) = 120*(t-10.85)
-200 t + 18110 = 120 t - 1302
320 t = 19412
t = 60.67 o
C
EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO
Msc. Widmar Aguilar
MAYO 2023
QPERDIDO = cm (∆ )
QPERDIDO = (C cal/g o
C) * 250 g*(21.32- 100) o
C
Qganado = cm (∆ )
Qganado = (1 cal/g o
C) *501.2 g*(21.32- 20) o
C
Se tiene: claton = 0.092 cal/g o
C
- QPERDIDO = QGANADO + QGANADO calorimetro
− (C cal/g o
C) * 250 g*(21.32- 100) o
C = (1 cal/g o
C) * 501.2 g*(21.32- 20) o
C + (0.092
cal/g o
C) * 200 g*(21.32- 20) o
C
− 250 > ∗ (−78.68) = 501.2 ∗ 1.32+ 18.4*1.32
19670 C = 661.58+24.29
C = 0.035 cal/g o
C
- QPERDIDO = QGANADO + QGANADO calorimetro
- 0.11*400*(t-100) = 1*500(t-20)+0.092*300*(t-20)+ 0.03 * 200* (t-25)+
-44*(t-100) = 500(t-20)+27.6(t-20)+6(t-25)
-44t+4400 =500t-10000+27.6t -552+6t-150
4400 = 577.6t -10702
577.6 t = 15102
= 26.2 o
C

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TEMPERATURA_CALOR_MAIZTEGUI.pdf

  • 1. EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO Msc. Widmar Aguilar MAYO 2023 TEMPERATURA Y CALOR
  • 2. EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO Msc. Widmar Aguilar MAYO 2023
  • 3. EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO Msc. Widmar Aguilar MAYO 2023
  • 4. EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO Msc. Widmar Aguilar MAYO 2023 Se tiene la relación: . = . = ; x-32 = 65.7 X = 32+65.7 X = 97.7 o F ( ) = = ; 32- x = 9 X = 32- 9
  • 5. EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO Msc. Widmar Aguilar MAYO 2023 X = 23 o F b) ( ) = = ; 32- x =54 X = 32- 54 X = -22 o F = = ; x =66.67 X = 66.67 o C
  • 6. EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO Msc. Widmar Aguilar MAYO 2023 = = ; x = - 3.89 X = -3.89 o C = - = ; x = - 17.78 X = - 17.78 o C − 0 100 = 8 − 32 212 − 32
  • 7. EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO Msc. Widmar Aguilar MAYO 2023 = ; x = - 13.33 X = -1 3.33 o C 0 − 100 = 32 − (−50) 212 − 32 - = ; x = - 45.55 X = -45.55 o C 0 − 100 = 32 − 212 − 32 - = ; -180 x = 3200 -100x
  • 8. EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO Msc. Widmar Aguilar MAYO 2023 -80 X = 3200 X= -40 X = -40 o C = - 40 o F 50 − 0 100 = − 273 373 − 273 = ; 50 = x-273 X = 323 o K Se sabe que: 0 o C = 273 o K 50 − 0 100 = − 273 373 − 273
  • 9. EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO Msc. Widmar Aguilar MAYO 2023 ( ) = ; = 273-X = 35 X = 238 o K 72 − 32 212 − 32 = − 273 373 − 273 = ; x-273 = 22.22 X = 295.22 o K 32 − 0 212 − 32 = 273 − 373 − 273 = ; 273 -x = 17.77
  • 10. EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO Msc. Widmar Aguilar MAYO 2023 X = 255.22 o K 32 − (−215) 212 − 32 = 273 − 373 − 273 = ; 273 -x = 137.22 X = 135.77 o K 0 − (−273) 100 − 0 = 32 − 212 − 32 = ; 32-x = 491.4 X = - 459.4 o F
  • 11. EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO Msc. Widmar Aguilar MAYO 2023 CALOR: Q = cm∆ C = 1 cal/g 0 C Q = 1 cal/g 0 C * 350 g* (85-35) 0 C Q = 350 * 50 cal Q = 17500 cal De: Q= c m ∆ 0.95 cal = c (1g)(1) o C c = 0.95 !"# Q= c m ∆ ; c = 0.032 cal /g*o C 500 cal = 0.032 * 50 * (t-50) t-50 = 312.5 t = 362.5 o C
  • 12. EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO Msc. Widmar Aguilar MAYO 2023 d = m/V Q= c m ∆ Q = c d V ∆ Qhierro = 0.11 *7.87*V*10 Qplatino = 0.032 *21.45*V*10 $%&'(() $*+,-&.) = 0.11 ∗7.87∗V∗10 0.032 ∗21.45∗V∗10 $%&'(() $*+,-&.) = 0.11 ∗7.87 0.032 ∗21.45 = 1.26 Qhierro = 1.26 Qplatino 195 o F = 90.55 o C 284 o K = 10.85 o C Qperdido = Q ganado M1 c1 (∆ ) = M2 c2 (∆ ) : - Qperdido = Q gana Qperdido = 200 g* 1 ca/g o C *(t- 90.55) 200 g* 1 ca/g o C *(90.55-t) = 120 g* 1 ca/g o C *(t-10.85) 200*( 90.55-t) = 120*(t-10.85) -200 t + 18110 = 120 t - 1302 320 t = 19412 t = 60.67 o C
  • 13. EJERCICIOS DE LA FISICA DE MAIZTEGUI SABATO Msc. Widmar Aguilar MAYO 2023 QPERDIDO = cm (∆ ) QPERDIDO = (C cal/g o C) * 250 g*(21.32- 100) o C Qganado = cm (∆ ) Qganado = (1 cal/g o C) *501.2 g*(21.32- 20) o C Se tiene: claton = 0.092 cal/g o C - QPERDIDO = QGANADO + QGANADO calorimetro − (C cal/g o C) * 250 g*(21.32- 100) o C = (1 cal/g o C) * 501.2 g*(21.32- 20) o C + (0.092 cal/g o C) * 200 g*(21.32- 20) o C − 250 > ∗ (−78.68) = 501.2 ∗ 1.32+ 18.4*1.32 19670 C = 661.58+24.29 C = 0.035 cal/g o C - QPERDIDO = QGANADO + QGANADO calorimetro - 0.11*400*(t-100) = 1*500(t-20)+0.092*300*(t-20)+ 0.03 * 200* (t-25)+ -44*(t-100) = 500(t-20)+27.6(t-20)+6(t-25) -44t+4400 =500t-10000+27.6t -552+6t-150 4400 = 577.6t -10702 577.6 t = 15102 = 26.2 o C