500 grams of ice at -10^oC will be melted down to become water at 5^oC. If the specific heat of ice is 0.5 cal/gr^oC, the heat of fusion for ice is 80 cal/gr and the specific heat of water is 1 cal/gr^oC. Determine the amount of heat needed * 20 poin
1. 500 grams of ice at -10^oC will be melted down to become water at 5^oC. If the specific heat of ice is 0.5 cal/gr^oC, the heat of fusion for ice is 80 cal/gr and the specific heat of water is 1 cal/gr^oC. Determine the amount of heat needed * 20 poin
Jawaban:
Penjelasan:
- heat absorbed to reach 0 degree C
[tex]Q_1=mc\Delta T=500 gram\times 0.5 cal/gramK\times (0-(-10))K=2500 cal[/tex]
- heat absorbed to make it water
[tex]Q_2=mL=500gram\times 80 cal/gram=40000 cal[/tex]
- heat absorber to reach 5 degree C of water
[tex]Q_3=mc\Delta T=500 gram times 1 cal/gramK\times (5-0)K=2500 cal[/tex]
total heat absorbed
[tex]Q=Q_1+Q_2+Q_3=2500cal+40000cal+2500cal=45000cal[/tex]
2. M-gram ice at 0ºC placed in 340 grams water at 20ºC in a specific container. Latent heat of fusion of ice (Lf) = 80 cal/g, the specific heat of water (cwater) = 1 cal/gºC. All the ice melts and the thermal equilibrium = 5ºC. Find mass of ice!
Massa dari es tersebut adalah 60 gr. Gunakan persamaan:
Q lepas = Q terima
Penjelasan dengan langkah-langkahDiketahui: M-gram es bersuhu 0ºC dimasukkan ke dalam 340 gram air bersuhu 20ºC dalam wadah tertentu. Panas laten peleburan es (Lf) = 80 kal/g, kalor jenis air (air) = 1 kal/gºC. Semua es mencair dan kesetimbangan termal = 5ºCDitanya: Temukan massa dari es tersebut!Jawab:Langkah 1
Gunakan persamaan:
Q lepas = Q terima
(m x c x ΔT) air = (m x L) es + (m es x c air x ΔT es)
Langkah 2
(340) (1) (20 - 5) = m (80) + m(1) (5 - 0)
(340) (15) = 85 m
5.100 = 85 m
m = 60 gr
Pelajari lebih lanjutMateri tentang kalor: https://brainly.co.id/tugas/52034962
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3. How many joules are energy required to melting 100g ice 0"C becomes water 50'C. Specific laten heat of fusion = 336.000 J/Kg Specific heat capacity of water = 4200 J/Kg "C
Jawaban:
energy required 54.600joule or 54,6KJ
Penjelasan:
cara difoto
4. ice starts to melt after it ....... heat from its surroundings and its temperature reaches ...... °C. this is called the ...... point of ice the temperature of ice remains at ....... °C until all the ice has completely changed into water.
Jawaban:
1. absorbs/receives
2. zero
3. melting
4. zero
5. how much heat is needed to melt 50 grams of ice at 0°C to water at 25°C
Δt = 25-0 => 25°C
the kalor of water is 4.200 Joule/kg°C
Q = m . c . Δt
Q = 0,05 kg . 4200 . 25
Q = 5250 joule
sorry if there is incorrect answer
6. An object with a mass of 200 g has a heat capacity of 104 J/oC. The specific heat of the substance is.
An object with a mass of 200 g has a heat capacity of 104 J/°C. The specific heat of the substance is 20.8 J/kg°C
Solution:
Known:
m = 200 gram = 0,2 kg
C = 104 J/°C
Asked:
c = ...?
Answer:
c = m × c
= 0,2 × 104
= 20.8 J/kg°C
So, the specific heat of the substance is 20.8 J/kg°C.
7. At 0 ° C the sublimation heat of the ice is 675.7 cal / g, while the heat of vaporization of the water is 595.9 cal / g. Calculating the rate of change of pressure steam temperature and ice water at 0 ° C. At this temperature the pressure of water vapor is 4.58 mm Hg.
Jawaban:
The rate of change of pressure with temperature for a substance in a given phase (solid, liquid, or gas) is determined by its vapor pressure curve. The vapor pressure curve shows the pressure of a substance's vapor in equilibrium with its condensed phase (solid or liquid) at a given temperature.
For water, the vapor pressure curve shows that the vapor pressure of water increases as the temperature increases. At 0 °C, the vapor pressure of water is 4.58 mm Hg. As the temperature increases, the vapor pressure of water will also increase, and the rate of change of pressure with temperature will be positive.
The sublimation heat of ice (675.7 cal/g) and the heat of vaporization of water (595.9 cal/g) are both related to the energy required to change the phase of water (from solid to vapor or liquid to vapor, respectively). These values are not directly related to the vapor pressure curve and do not affect the rate of change of pressure with temperature at a given temperature.
Jawaban:
= -37.5 atm/s
Penjelasan:
To calculate the rate of change of pressure for steam and ice water at 0°C, we can use the rate of change of pressure equation, which is defined as the change in pressure per unit time. This equation can be written as follows:
ΔP/Δt = hL(Pv - P)
where:
ΔP is the change in pressure
Δt is the change in time
hL is the rate of change of pressure, which is a constant that depends on the temperature and density of the fluid
Pv is the vapor pressure of the fluid
P is the current pressure of the fluid
To calculate the rate of change of pressure for steam at 0°C, we can use the values given in the question, which are:
hL(steam) = (595.9 cal/g) / (18 g/mol) = 33.1 cal/mol.atm
Pv(steam) = 4.58 mm Hg = 0.006 atm
P(steam) = 1 atm (standard atmospheric pressure at 0°C)
Using the above equation, we can calculate the rate of change of pressure for steam at 0°C as follows:
ΔP/Δt = hL(Pv - P)
= (33.1 cal/mol.atm) * (0.006 atm - 1 atm)
= -0.198 atm/s
On the other hand, to calculate the rate of change of pressure for ice water at 0°C, we can use the values given in the question, which are:
hL(ice water) = (675.7 cal/g) / (18 g/mol) = 37.5 cal/mol.atm
Pv(ice water) = 0 atm (the vapor pressure of ice is 0 at 0°C)
P(ice water) = 1 atm (standard atmospheric pressure at 0°C)
Using the above equation, we can calculate the rate of change of pressure for ice water at 0°C as follows:
ΔP/Δt = hL(Pv - P)
= (37.5 cal/mol.atm) * (0 atm - 1 atm)
= -37.5 atm/s
8. an example of conductor of heat is.........
an example of conductor of heat is iron
An example of conductor of heat is copper9. Ice with a mass of 400 grams at a temperature of 0 oC will be melted into 400 grams of water at 0 oC, then the amount of heat absorbed is …. (Les = 80 Cal/gram) * 30 poin
Jawaban:
Penjelasan:
the heat absorbed
[tex]Q=mL\\Q=400g\times 80cal/gram=32000 cal[/tex]
10. Jenna buys.. A. a bowl of ice cream B. a glass of ice cream C. a scoop of ice cream D. a cube of ice cream
C. a scoop of ice cream
knp c krn es krimkan pake skop es krim, jd kalau mau beli mbak/mas satu skop es krim ya. gak mungkin segelas, semangkuk, sekubus.
11. An object with a mass of 200 g has a heat capacity of 104 J/oC. The specific heat of the substance is
Jawaban:
520 J/Kg C
diketahui :
m = 200 = 0,2
C = 104 J/C
ditanya :
c = ?
dijawab :
c = C/m
c = 104/0,2
c = 520 J/Kg C
12. the amount of the heat needed for 2 kg of ice at -5 °c if it is heated until all the ice melts is?
Dik : Massa : 2kg
ΔT = 0 -(-5) = 5°C
Ces = 2100 (satuannya lupa)
L = 336.000
Q1 = m.Ces.ΔT
= 2 × 2100 × 5
= 21000 J
Q2 = m × L
= 2 × 336.000
= 672.000 J
Qtot = Q1 + Q2
= 21.000 + 672.000
= 693.000 J
Sekian.
13. 1. (-) ....... a. Rudy drank not a glass of ice teab. Rudy didn't drink a glass of ice teac. Rudy didn't drank a glass of ice tead. Rudy not drinking a glass of ice tea2. (?) ...............a. Did Rudy drink a glass of ice tea?b. Did Rudy drank a glass of ice tea?c. Did Rudy drinked a glass of ice tea?d. Did Rudy drinking a glass of ice tea?tolong di jawab yang benar ya
1. (-) Rudy didn’t drink a glass of ice tea (B)
2. (?) Did Rudy drink a glass of ice tea? (A)
14. A block of 0°C ice with the mass of 50 kg mass slides on a horizontal surface. The initial velocity of the ice is 6.0 m/s and it stops after a distance of 28.3 m. How much ice melts due to the friction? (specific latent heat of fusion of ice, Lf = 80 cal/g and 1 cal = 4.18 J and ignore the heat transfers to the environment) ... (A) 2.7 g (B) 4.7 g (C) 11.2 g (D) 47.0 g (E) 57.2 g
D) 47.0 g
Penjelasan:
Jumlah pertama ditambahkan seluruh nya
15. What happens when a piece of ice melts? A. the piece of ice becomes smaller. B. the piece of ice changes into water. C. the piece of ice takes in heat D. The temperature of the air around the ice falls
Jawaban:
E. All the above
karna semua jawaban benar
16. a block of ice which has 20 kg of mass is receiving 67 kJ of heat. The number of ice melts is... (the melting point of ice is 335 kJ/kg)
Q = m.L
67 = m.335
m = 67:335
m = 0,2 kg
17. Exercise:1.How much heat is required to melt 10 kg of alcohol in solid? The heat of fusion ofalcohol is 6,9 x 104 J kg-1.2.How much heat is required to vaporize 5 kg of water? The heat of vaporization of wateris 2,27 3 106 J kg-1.
Jawaban:
1. 690000 joule = 690 kj
2. 11365 kj
Penjelasan:
1. diket : m = 10 kg
U = 6,9 x 10 pngkt 4 joule/ kg
jawab : Q = m x U = 10 kg x 69000 j/kg
= 690000 joule
2. diket : m = 5 kg
U = 2.273.000 j/ kg
jawab : Q = m x U = 5 kg x 2.273.000 j/kg
= 11.365.000 joule
18. the amount of heat needed for 2 kg of ice at -5 celcius if it is heated until all ice melts is .... (c ice = 2.100 J/kg celcius L=336.000j/kg)
Q= m × L Q= 2 × 336.000 Q= 672.000 Are the temperature of ice just -5°C?Q = m.c.T + m.L
= 2 kg. 2100. 5 + 2.336000
= 21000 + 672000
= 693000 J
19. 5 cups of ice-cream and 4 cups of milk tea cost $26.80 whereas 7 cups of ice-cream and 6 cups of milk tea cost $38.60.find the the cost of 8 cup of ice-cream and 5 cups of milk tea.
Jawab:
Penjelasan dengan langkah-langkah:
20. An object with a mass of 200 g has a heat capacity of 104 J/oC. The specific heat of the substance is
An object with a mass of 200 g has a heat capacity of 104 J/°C. The specific heat of the substance is 20.8J/kg°C
Solution:Known:
m = 200 gram = 0,2 kg
C = 104 J/°C
Asked:
c = ...?
Answer:
c = m × c
= 0,2 × 104
= 20.8J/kg°C
So, the specific heat of the substance is 20.8 J/kg°C.
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Oktober 28, 2022
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