Температура и топлина – проблеми и решенија
1. On a thermometer X, the freezing point of water at -30o and the boiling point of water at 90o. 60OX = … oC.
Познато:
The freezing point of water = -30o
The boiling point of water = 90o
Барани: 60oX = … oC
Решение:
On the Fahrenheit scale, the freezing point of water is 32oF and the boiling point of water is 212oF. Between the freezing point and the boiling point, 212o - 32o = 180o.
On the Celsius scale, the freezing point of water is 0oC and the boiling point of water is 100oC. Between the freezing point and the boiling point, 100o - 0o = 100o.
On the X scale, the freezing point of water is -30oX and the boiling point of water is 90oX. Between the freezing point and the boiling point, 90o – (-30o) = 90o + 30o = 120o.
Change the X scale to the Celsius scale :
2. A metal rod heated from 30oГолемото Ц 80oC. The final length of the rod is 115 см. Коефициентот на линеарна експанзија is 3.10-3 oC-1. What is the initial length of the metal rod?
Познато:
The initial temperature (T1) = 30oC
Конечната температура (T2) = 80oC
Промената на температурата (ΔT) = 80oЦ - 30oC = 50oC
Коефициент на линеарна експанзија (α) = 3.10-3 oC-1
The final length of the metal (L) = 115 cm
Барани: The initial length of the metal rod (Lo)
Решение:
The equation of the linear expansion :
L = Lo + ΔL
L = Lo + α Lo ΔT
L = Lo (1 + α ΔT)
115 = Лo (1 + 3.10)-3.50)
115 = Лo (1 + 150.10)-3)
115 = Лo (1 + 0.15)
115 = Лo (1.15)
Lo = 115/1.15
Lo = 100 см
3. The initial length of a brass rod is 40 см. After heated, the final length of the brass is 40.04 cm and the final temperature is 80oC. If the coefficient of linear expansion of the brass is X 2.0 10-5 oC-1, what is the initial temperature of the brass rod.
Познато:
Конечната температура (T2) = 80oC
Почетната должина (Lo) = 40 см
The final length (L) = 40.04 cm
The increase in length (ΔL) = 40.04 cm – 40 cm = 0.04 cm
Коефициент на линеарна експанзија (α) = 2.0 x 10-5 oC-1
Решение: Почетна температура (T1)
Solution ;
The equation of the linear expansion :
L = Lo + α Lo ΔT
Л – Лo = α Lo ΔT
ΔL = αLo ΔT
ΔL = αLo (T2 - Т.1)
0.04 = (2.0 x 10-5)(40)(80 – Т1)
0..04 = (80 x 10-5)(80 – Т1)
0.04 = 0.0008 (80 – Т1)
0.04 = 0.064 – 0.0008 Т1
0..0008 T1 = 0.064 - 0.040
0.0008 Т1 = 0.024
T1 = 30oC
4. Two metal rods with the same size but different type, as shown in figure below. The thermal conductivity of metal I = 4 times the thermal conductivity of metal II. What is the temperature between both metals.
Познато:
The size of both rods is the same.
The thermal conductivity of metal I = 4k
The thermal conductivity of metal II = k
The temperature of the one end of metal I = 500 C
The temperature of the one end of metal II = 00 C
Сакаше: The temperature between both the metal rods
Решение:
Равенката на топлинска спроводливост:
Q/t = the rate of heat conduction, k = топлинска спроводливост, А = површината на попречниот пресек, Т1-T2 = промената на температурата, л = the length of the rod.
The temperature between both rods :
The temperature at the center between both the metals rods is 40oC.
5.
(1) Conductivity of metal
(2) The difference of temperature
(3) The length of metal
(4) Mass of metal
Factors that determine the rate of heat conduction on metals are…..
Решение:
Based on the equation of heat conduction, factors that determine the rate of heat conduction on metals are conductivity of metal (k), the difference of temperature (T) and the length of metal (l).
6. Two rods of the same size but different type, as shown in the figure below. The thermal conductivity of rod P is 2 times the thermal conductivity of rod Q. What is the temperature between both rods.
Познато:
Both rods have the same size.
Thermal conductivity of rod P (kP) = 2 илјади
The thermal conductivity of rod Q (kQ) = k
Сакаше: The temperature between both rods
Решение:
Равенката на топлинска спроводливост:
Q/t = the rate of heat conduction, k = топлинска спроводливост, А = површината на попречниот пресек, Т1-T2 = промената на температурата, л = the length of the rod.
The temperature at the center between both rods :
8. 100-gram oil at 20oC and 50-gram iron at 75 oC are placed in 200-gram iron container. The increase in temperature of the container is 5oC and the specific heat of oil is 0.43 cal/g oC. What is the specific heat of the iron?
Познато:
Mass of iron container (m) = 200 gr
The initial temperature of the iron container (T1) = the temperature of oil = 20oC
The final temperature of the iron container (T2) = 20oC+5oC = 25oC
Mass of oil (m) = 100 gram
The specific heat of oil (cмасло) = 0.43 cal/g oC
The initial temperature of oil (T1) = 20oC
The final temperature of oil (T2) = 20oC+5oC = 25oC
Mass of iron (m) = 50 gram
The initial temperature of oil (T1) = 75oC
The final temperature of oil (T2) = 25oC
Барани: The specific heat of iron (c iron)
Решение:
Heat released by iron :
Q = m c ΔT = (50)(c)(75-25) = (50)(c)(50) = 2500c calorie
Heat absorbed by the iron container :
Q = m c ΔT = (200)(c)(25-20) = (200)(c)(5) = 1000c calorie
Heat absorbed by oil :
Q = m c ΔT = (100)(0.43)(25-20) = (43)(5) = 215 calorie
Black principle states that in a isolated system, heat released by the hotter object, absorbed by the cooler object.
Q release = Q absorb
2500c = 1000c + 215
2500c – 1000c = 215
1500c = 215
c = 215/1500
c = 0.143 cal/g oC
9. A 200-gram water at 20°C placed in 50-gram ice at -2°C. If the change of heat just between water and ice, what is the final temperature of the mixture? The specific heat of water is 1 cal/gr°C, the specific heat of ice is 0.5 cal/gr°C, the heat of fusion for ice is 80 cal/gr.
Познато:
Mass of water (mвода) = 200 грама
Temperature of water (Tвода) = 20oC
The specific heat of water (cвода) = 1 cal/gr°C
Mass of ice (mмраз) = 50 грама
Температура на мразот (Tмраз) = -2oC
The specific heat of ice (cмраз) = 0.5 cal/gr°C
The heat of fusion for ice (L) = 80 cal/gr
Решение:
Heat to increases ice from -2oC до 0oВ:
Q = mc ΔT
Q = (50 gram)(0.5 cal/gr°C)(0oC – (-2oВ))
Q = (50)(0.5 cal)(2)
Q = 50 calorie
Heat for melting all ice :
Q = m L = (50 gram)(80 cal/gram) = 4000 calorie
Heat for decrease temperature of all water from 20oC до 0oВ:
Q = mc ΔT
Q = (200 gram)(1 cal/gr°C)(0oC – (20oВ))
Q = (200)(1 cal)(-20)
Q = -4000 calorie
Plus sign indicates that the heat is added, the minus sign indicates that heat released.
50-calorie of heat needed to increase the temperature of ice to 0oC and 4000-calorie needed to melting all ice. Total heat = 4050 calorie. The heat released by water is 4000 calorie.
Some of the ice not melting, so the final temperature of ice and water is 0oC.
10. A 200-gram aluminum at 20oC placed in 100-gram water at 80oC in a container. The specific heat of aluminum is 0.22 cal/g oC and the specific heat of water is 1 cal/g oC. What is the final temperature of aluminum?
Познато:
Mass of aluminum = 200 gram
Temperature of aluminum = 20oC
Mass of water = 100 gram
Temperature of water = 80oC
The specific heat of aluminum = 0.22 cal/g oC
The specific heat of water = 1 cal/g oC
Сакаше: The final temperature of aluminum
Решение:
Aluminum and water in thermal equilibrium so that the final temperature of aluminum = the final temperature of water.
Heat released by hot water (Q release) = heat absorbed by aluminum (Q absorb)
mвода в (ΔT) = mалуминиум в (ΔT)
(100)(1)(80 – Т) = (200)(0.22)(Т – 20)
(100)(80 – Т) = (44)(Т – 20)
8000 – 100T = 44T – 880
8000 + 880 = 44T + 100T
8880 = 144T
Т = 62oC
11. A 50-gram metal at 85 °C placed in 50 gram water at 29.8 °C. The specific heat of water = 1 cal.g -1 .°C-1. The final temperature is 37 °C. What is the specific heat of metal.
Познато:
Mass of metal (mметал) = 50 грама
Temperature of metal = 85oC
Mass of water (mвода) = 50 грама
Temperature of water = 29,8oC
The specific heat of water (cвода) = 1 cal.g -1 .°C-1
The final temperature of water = 37oC
Барани: The specific heat of metal (c metal)
Решение:
Heat released by hot metal (Q release) = heat absorbed by water (Q absorb)
mметал в (ΔT) = mвода в (ΔT)
(50)(в)(85 – 37) = (50)(1)(37 – 29.8)
(в)(85 – 37) = (1)(37 – 29.8)
48 c = 7.2
c = 0.15 cal.g -1 .°C-1
12. A block of ice with mass of 50-gram at 0°C and 200-gram water at 30°C, placed in a container. . If the specific heat of water is 1 cal.g- 1 ° C -1 and the heat of fusion for ice is 80 cal.g -1. What is the final temperature of the mixture.
Познато:
Mass of ice (mмраз) = 50 грама
The temperature of ice = 0°C
Mass of water (mвода) = 200 грама
Temperature of water = 30oC
The specific heat of water (cвода) = 1 cal.g- 1 ° C -1
The heat of fusion for ice (Lмраз) = 80 cal.g -1
Сакаше: Конечната температура
Решение:
Estimate the final condition :
Heat released by water to decrease its temperature from 30oC до 0oВ:
Qпорака = мвода cвода (ΔT) = (200)(1)(30-0) = (200)(30) = 6000
Heat needed to melting all ice :
Q = mмраз Lмраз= (50)(80) = 4000
Heat used to melting all ice is 4000, while heat released by water is 6000. Can be concluded that the final temperature of the mixture above 0oC.
Black principle :
Heat released by water = heat for melting all ice + heat to increase the temperature of ice.
(mвода)(в)вода)(ΔT) = (mмраз)(Лмраз) + (ммраз)(в)вода)(ΔT)
(200)(1)(30-Т) = (50)(80) + (50)(1)(Т-0)
(200)(30-Т) = (50)(80) + (50)(Т-0)
6000 – 200T = 4000 + 50T – 0
6000 – 4000 = 50T + 200T
2000 = 250T
Т = 2000/250
Т = 8oC
