Temperature and Heat calculations Flashcards
What is 37°C in Kelvin?
To convert Celsius to Kelvin, use T_K = T_C + 273.15. Here, T_C = 37°C. So, T_K = 37 + 273.15 = 310.15 K.
A steel rod is 2 m long at 20°C. How much does it expand when heated to 50°C? (α_steel = 12 * 10^-6 /°C)
Use ΔL = α * L_0 * ΔT. L_0 = 2 m, α = 12 * 10^-6 /°C, ΔT = 50 - 20 = 30°C. ΔL = (12 * 10^-6) * 2 * 30 = 0.000012 * 60 = 0.00072 m = 0.72 mm.
How much heat is needed to raise the temperature of 0.5 kg of water from 25°C to 75°C? (c_water = 4186 J/kg·K)
Use Q = m * c * ΔT. m = 0.5 kg, c = 4186 J/kg·K, ΔT = 75 - 25 = 50°C (same as 50 K). Q = 0.5 * 4186 * 50 = 104650 J = 104.65 kJ.
A 0.1 m^2 copper plate conducts heat between 100°C and 20°C over a thickness of 0.01 m. What is the heat current? (k_copper = 401 W/m·K)
Use H = k * A * (T_H - T_C) / L. k = 401 W/m·K, A = 0.1 m^2, T_H - T_C = 100 - 20 = 80°C, L = 0.01 m. H = 401 * 0.1 * 80 / 0.01 = 401 * 8 * 100 = 320800 W = 320.8 kW.
How much heat is required to melt 2 kg of ice at 0°C? (L_f_ice = 3.34 * 10^5 J/kg)
Use Q = m * L_f. m = 2 kg, L_f = 3.34 * 10^5 J/kg. Q = 2 * 3.34 * 10^5 = 6.68 * 10^5 J = 668 kJ.
A 0.2 kg aluminum block at 100°C is dropped into 1 kg of water at 20°C. What is the final temperature? (c_al = 900 J/kg·K, c_water = 4186 J/kg·K)
Use calorimetry: Q_lost = Q_gained, so m_al * c_al * (T_i_al - T_f) = m_w * c_w * (T_f - T_i_w). m_al = 0.2 kg,
c_al = 900 J/kg·K,
T_i_al = 100°C,
m_w = 1 kg,
c_w = 4186 J/kg·K,
T_i_w = 20°C.
Set up:
0.2 * 900 * (100 - T_f) = 1 * 4186 * (T_f - 20).
Simplify: 180 * (100 - T_f) = 4186 * (T_f - 20).
18000 - 180 * T_f = 4186 * T_f - 83720. Combine:
18000 + 83720 = 4186 * T_f + 180 * T_f. 101720 = 4366 * T_f. T_f = 101720 / 4366
≈ 23.3°C.
A 0.05 m^2 wall has two layers: 0.02 m wood (k_wood = 0.1 W/m·K) and 0.01 m insulation (k_ins = 0.04 W/m·K). If the temperature difference is 20°C, what is the heat current?
Use thermal resistance in series: R_total = R_1 + R_2, where R = L / (k * A). For wood: R_1 = 0.02 / (0.1 * 0.05) = 0.02 / 0.005 = 4 K/W. For insulation: R_2 = 0.01 / (0.04 * 0.05) = 0.01 / 0.002 = 5 K/W. R_total = 4 + 5 = 9 K/W. H = (T_H - T_C) / R_total = 20 / 9 ≈ 2.22 W.
A 0.3 kg iron sphere (e = 0.9, radius = 0.05 m) at 500 K radiates heat. What is its heat loss rate? (σ = 5.67 * 10^-8 W/m^2·K^4)
Use H = σ * e * A * T^4. Area A = 4 * π * r^2, r = 0.05 m, so A = 4 * 3.1416 * 0.05^2 = 4 * 3.1416 * 0.0025 = 0.031416 m^2. σ = 5.67 * 10^-8 W/m^2·K^4, e = 0.9, T = 500 K. H = 5.67 * 10^-8 * 0.9 * 0.031416 * 500^4. First, 500^4 = 62500000000. Then, 5.67 * 10^-8 * 0.9 = 5.103 * 10^-8. Next, 5.103 * 10^-8 * 0.031416 = 1.603 * 10^-9. Finally, 1.603 * 10^-9 * 62500000000 = 100.19 W ≈ 100 W.
How much heat is needed to convert 0.1 kg of ice at -10°C to water at 50°C? (c_ice = 2090 J/kg·K, L_f_ice = 3.34 * 10^5 J/kg, c_water = 4186 J/kg·K)
Three steps: (1) Heat ice to 0°C: Q_1 = m * c_ice * ΔT = 0.1 * 2090 * (0 - (-10)) = 0.1 * 2090 * 10 = 2090 J. (2) Melt ice: Q_2 = m * L_f = 0.1 * 3.34 * 10^5 = 33400 J. (3) Heat water to 50°C: Q_3 = m * c_water * ΔT = 0.1 * 4186 * (50 - 0) = .Concurrent 0.1 * 4186 * 50 = 20930 J. Total Q = Q_1 + Q_2 + Q_3 = 2090 + 33400 + 20930 = 56420 J = 56.42 kJ.
A 0.200 kg aluminum container holds 0.300 kg of water at 20 degrees C. If 0.500 kg of iron at 150 degrees C is added, what is the final temperature? No heat is lost to surroundings.
Data: Specific heat of aluminum = 900 J/kg K, water = 4190 J/kg K, iron = 450 J/kg K
31.8 degrees C
Question: 0.050 kg of ice at 0 degrees C is added to 0.200 kg of water at 25 degrees C in a 0.100 kg copper container. What is the final temperature?
Data: Specific heat of copper = 390 J/kg K, water = 4190 J/kg K, latent heat of fusion = 334000 J/kg
4.8 degrees C
Question: A 0.300 m brass rod and a 0.700 m copper rod are joined end-to-end. The brass end is at 100 degrees C, the copper end at 0 degrees C. Cross-sectional area is 0.004 square meters. What is the junction temperature?
Data: Conductivity of brass = 109 W/m K, copper = 385 W/m K
57.6 degrees C
Question: How much ice melts in 10 minutes due to heat conducted through a 0.500 m copper rod with ends at 100 degrees C and 0 degrees C? Area = 0.006 square meters.
Data: Conductivity of copper = 385 W/m K, latent heat of fusion = 334000 J/kg
Answer: 0.83 kg
Question: A 0.150 kg steel container at 20 degrees C holds 0.100 kg of water and 0.020 kg of ice at 0 degrees C. Add 0.400 kg of lead at 200 degrees C. What is the final temperature?
Data: Specific heat of steel = 500 J/kg K, water = 4190 J/kg K, lead = 130 J/kg K, latent heat of fusion = 334000 J/kg
Answer: 23.5 degrees C
Question: A 0.600 m aluminum rod conducts heat from 80 degrees C to 10 degrees C. Area = 0.003 square meters. How much heat flows in 2 minutes?
Data: Conductivity of aluminum = 205 W/m K
8,610 J
Question: A cylindrical container, radius 0.050 m and height 0.200 m, at 10 K radiates to walls at 50 K. Emissivity = 0.150. How much mass vaporizes per hour? Heat of vaporization = 15000 J/kg
Answer: 4.92 g
Question: 0.080 kg of ice at 0 degrees C melts in 0.250 kg of water at 40 degrees C in a 0.120 kg brass container. What is the final temperature?
Data: Specific heat of brass = 109 J/kg K, water = 4190 J/kg K, latent heat of fusion = 334000 J/kg
Answer: 22.7 degrees C
Question: A 0.400 m copper rod and 0.300 m steel rod are joined. Copper end at 90 degrees C, steel end at 20 degrees C. Area = 0.002 square meters. What is the junction temperature?
Data: Conductivity of copper = 385 W/m K, steel = 50 W/m K
Answer: 84.8 degrees C
Question: How much heat does a 0.250 kg copper block lose cooling from 120 degrees C to 30 degrees C?
Data: Specific heat of copper = 390 J/kg K
Answer: 8775 J
Question: A 0.100 kg iron container with 0.200 kg water at 15 degrees C receives 0.300 kg aluminum at 180 degrees C. What is the final temperature?
Data: Specific heat of iron = 450 J/kg K, water = 4190 J/kg K, aluminum = 900 J/kg K
53.6°C
Question: A 0.500 m brass rod conducts heat from 95 degrees C to 5 degrees C. Area = 0.007 square meters. What is the heat current?
Data: Conductivity of brass = 109 W/m K
Answer: 137.3 W
Question: A container at 5 K, radius 0.030 m and height 0.150 m, radiates to 60 K walls. Emissivity = 0.250. How much mass is lost per hour? Heat of vaporization = 20000 J/kg
1.12 kg
Question: 0.030 kg ice at 0 degrees C and 0.150 kg water at 30 degrees C in a 0.090 kg steel container. What is the final temperature?
Data: Specific heat of steel = 500 J/kg K, water = 4190 J/kg K, latent heat of fusion = 334000 J/kg
Answer: 17.8 degrees C
Question: A 0.200 m copper rod and 0.400 m brass rod are joined. Copper at 100 degrees C, brass at 0 degrees C. Area = 0.005 square meters. What is the heat current?
Data: Conductivity of copper = 385 W/m K, brass = 109 W/m K
Answer: 128.3 W
Question: 0.600 kg lead at 300 degrees C is added to 0.200 kg water at 20 degrees C in a 0.150 kg aluminum container. What is the final temperature?
Data: Specific heat of lead = 130 J/kg K, water = 4190 J/kg K, aluminum = 900 J/kg K
Answer: 38.9 degrees C
Question: How much ice melts in 15 minutes by a 0.300 m steel rod conducting heat from 80 degrees C to 0 degrees C? Area = 0.004 square meters.
Data: Conductivity of steel = 50 W/m K, latent heat of fusion = 334000 J/kg
Answer: 0.054 kg
Question: A 0.100 kg brass container with 0.120 kg water at 25 degrees C receives 0.200 kg steel at 160 degrees C. What is the final temperature?
Data: Specific heat of brass = 109 J/kg K, water = 4190 J/kg K, steel = 500 J/kg K
Answer: 47.2 degrees C
Question: A 0.800 m aluminum rod conducts heat from 70 degrees C to 10 degrees C. Area = 0.006 square meters. How much heat flows in 5 minutes?
Data: Conductivity of aluminum = 205 W/m K
Answer: 46125 J
Question: A cylindrical container, radius 0.040 m and height 0.180 m, at 8 K radiates to 40 K walls. Emissivity = 0.180. How much mass is lost per hour? Heat of vaporization = 18000 J/kg
Answer: 2.65 g
A steel rod has a length of 2.00 m at 20°C. What is its new length at 80°C? (alpha_steel = 1.2 x 10^-5 °C^-1)
delta_L = L alpha delta_T = 2.00 (1.2 x 10^-5) (80 - 20) = 0.00144 m. New length = 2.00 + 0.00144 = 2.00144 m.
A copper wire (diameter 1.50 mm) is cooled from 25°C to -50°C. What’s its new diameter? (alpha_copper = 1.7 x 10^-5 °C^-1)
delta_D = D alpha delta_T = 1.50 (1.7 x 10^-5) (-50 - 25) = -0.0019125 mm. New diameter = 1.50 - 0.0019125 = 1.4981 mm.
A brass ring has an inner diameter of 5.000 cm at 15°C. At what temperature does it shrink to 4.995 cm? (alpha_brass = 1.9 x 10^-5 °C^-1)
delta_D = D alpha delta_T. -0.005 = 5.000 (1.9 x 10^-5) delta_T. delta_T = -0.005 / (9.5 x 10^-5) = -52.63 °C. T = 15 - 52.63 = -37.63°C.
An aluminum plate (length 0.50 m) is heated from 10°C to 100°C. What’s the length increase? (alpha_aluminum = 2.4 x 10^-5 °C^-1)
delta_L = L alpha delta_T = 0.50 (2.4 x 10^-5) (100 - 10) = 0.00108 m = 1.08 mm.
An iron bolt (diameter 10.00 mm) at 30°C must fit a 9.98 mm hole. At what temperature does it fit? (alpha_iron = 1.1 x 10^-5 °C^-1)
delta_D = D alpha delta_T. -0.02 = 10.00 (1.1 x 10^-5) delta_T. delta_T = -0.02 / (1.1 x 10^-4) = -181.82 °C. T = 30 - 181.82 = -151.82°C.
0.5 kg of wax at 20°C is heated to 70°C to melt it. Calculate the heat needed. (T_melt = 60°C, c_solid = 900 J/kg·K, c_liquid = 1200 J/kg·K, L_f = 150 x 10^3 J/kg)
Q1 = m c_solid delta_T = 0.5 900 (60 - 20) = 18000 J. Q2 = m L_f = 0.5 150000 = 75000 J. Q3 = m c_liquid delta_T = 0.5 1200 (70 - 60) = 6000 J. Q_total = 18000 + 75000 + 6000 = 99000 J.
How much heat melts 2 kg of lead from 25°C to 350°C? (T_melt = 327°C, c_solid = 130 J/kg·K, c_liquid = 150 J/kg·K, L_f = 25 x 10^3 J/kg)
Q1 = 2 130 (327 - 25) = 78520 J. Q2 = 2 25000 = 50000 J. Q3 = 2 150 (350 - 327) = 6900 J. Q_total = 78520 + 50000 + 6900 = 135420 J.
A copper rod (length 1.20 m) at 22°C must shrink to 1.199 m. What temperature is needed? (alpha_copper = 1.7 x 10^-5 °C^-1)
delta_L = L alpha delta_T. -0.001 = 1.20 (1.7 x 10^-5) delta_T. delta_T = -0.001 / (2.04 x 10^-5) = -49.02 °C. T = 22 - 49.02 = -27.02°C.
Heat 1 kg of ice from -10°C to 10°C. How much heat? (c_ice = 2100 J/kg·K, L_f = 334 x 10^3 J/kg, c_water = 4186 J/kg·K)
Q1 = 1 2100 10 = 21000 J. Q2 = 1 334000 = 334000 J. Q3 = 1 4186 10 = 41860 J. Q_total = 21000 + 334000 + 41860 = 396860 J.
A glass rod (length 0.80 m) at 18°C expands by 0.0005 m when heated. What’s the final temperature? (alpha_glass = 0.9 x 10^-5 °C^-1)
delta_L = L alpha delta_T. 0.0005 = 0.80 (0.9 x 10^-5) delta_T. delta_T = 0.0005 / (7.2 x 10^-6) = 69.44 °C. T = 18 + 69.44 = 87.44°C.
Heat 0.2 kg of nylon from 30°C to 230°C to melt it. How much heat? (T_melt = 220°C, c_solid = 1700 J/kg·K, c_liquid = 2000 J/kg·K, L_f = 160 x 10^3 J/kg)
Q1 = 0.2 1700 (220 - 30) = 64600 J. Q2 = 0.2 160000 = 32000 J. Q3 = 0.2 2000 (230 - 220) = 4000 J. Q_total = 64600 + 32000 + 4000 = 100600 J.
A steel plate (length 3.00 m) at 15°C reaches what temperature when its length is 3.002 m? (alpha_steel = 1.2 x 10^-5 °C^-1)
delta_L = L alpha delta_T. 0.002 = 3.00 (1.2 x 10^-5) delta_T. delta_T = 0.002 / (3.6 x 10^-5) = 55.56 °C. T = 15 + 55.56 = 70.56°C.
Heat 0.8 kg of paraffin from 25°C to 80°C. Calculate heat. (T_melt = 55°C, c_solid = 2200 J/kg·K, c_liquid = 2500 J/kg·K, L_f = 180 x 10^3 J/kg)
Q1 = 0.8 2200 (55 - 25) = 52800 J. Q2 = 0.8 180000 = 144000 J. Q3 = 0.8 2500 (80 - 55) = 50000 J. Q_total = 52800 + 144000 + 50000 = 246800 J.
An aluminum rivet (diameter 6.00 mm) at -50°C is heated until its diameter is 6.015 mm. What’s the temperature? (alpha_aluminum = 2.4 x 10^-5 °C^-1)
delta_D = D alpha delta_T. 0.015 = 6.00 (2.4 x 10^-5) delta_T. delta_T = 0.015 / (1.44 x 10^-4) = 104.17 °C. T = -50 + 104.17 = 54.17°C.
Heat 1.5 kg of tin from 20°C to 260°C to melt it. How much heat? (T_melt = 232°C, c_solid = 227 J/kg·K, c_liquid = 243 J/kg·K, L_f = 59 x 10^3 J/kg)
Q1 = 1.5 227 (232 - 20) = 72114 J. Q2 = 1.5 59000 = 88500 J. Q3 = 1.5 243 (260 - 232) = 10206 J. Q_total = 72114 + 88500 + 10206 = 170820 J.
