Articles
Register
Sign In
Search
souacarol
@souacarol
October 2020
1
324
Report
questão de física sobre calorimetria (o gabarito se encontra a frente)
Please enter comments
Please enter your name.
Please enter the correct email address.
Agree to
terms and service
You must agree before submitting.
Send
Lista de comentários
GFerraz
M = 100 g
a)
No estado líquido:
ΔQ = 2000 cal
ΔT = 140-60 = 80ºC
c = Q/m.ΔT
c = 2000/100.80
c = 2/8
c = 0,25 cal/gºC
b)
ΔQ = 5200 - 2000 = 3.200 cal
Q = m.L
L = Q/m
L = 3200/100
L = 32 cal/g
c)
ΔQ = 6400 - 5200 = 1200 cal
ΔT = 60ºC
Q = m.c.ΔT
1200 = 100.60.c
1200 = 6000.c
c = 1200/6000
c = 12/60
c = 0,2 cal/gºC
3 votes
Thanks 3
More Questions From This User
See All
souacarol
April 2022 | 0 Respostas
Responda
souacarol
October 2020 | 0 Respostas
Responda
souacarol
October 2020 | 0 Respostas
Responda
souacarol
October 2020 | 0 Respostas
Responda
souacarol
October 2020 | 0 Respostas
Responda
souacarol
October 2020 | 0 Respostas
Responda
souacarol
October 2020 | 0 Respostas
Responda
souacarol
October 2020 | 0 Respostas
Responda
souacarol
October 2020 | 0 Respostas
Responda
souacarol
October 2020 | 0 Respostas
Responda
×
Report "questão de física sobre calorimetria (o gabarito se encontra a frente).... Pergunta de ideia de souacarol"
Your name
Email
Reason
-Select Reason-
Pornographic
Defamatory
Illegal/Unlawful
Spam
Other Terms Of Service Violation
File a copyright complaint
Description
Helpful Links
Sobre nós
Política de Privacidade
Termos e Condições
direito autoral
Contate-Nos
Helpful Social
Get monthly updates
Submit
Copyright © 2024 ELIBRARY.TIPS - All rights reserved.
Lista de comentários
a)
No estado líquido:
ΔQ = 2000 cal
ΔT = 140-60 = 80ºC
c = Q/m.ΔT
c = 2000/100.80
c = 2/8
c = 0,25 cal/gºC
b)
ΔQ = 5200 - 2000 = 3.200 cal
Q = m.L
L = Q/m
L = 3200/100
L = 32 cal/g
c)
ΔQ = 6400 - 5200 = 1200 cal
ΔT = 60ºC
Q = m.c.ΔT
1200 = 100.60.c
1200 = 6000.c
c = 1200/6000
c = 12/60
c = 0,2 cal/gºC