This is the term for heat that is transferred molecule to molecule. For example from one air molecule to another in the atmosphere.单项选择题
A
Conduction
B
Adiabatic
C
Radiation
D
Solar
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Question text1. A shallow circular pond of radius [math: r=34m] freezes completely during very cold nights. The pond is in a geothermal area in which the temperature of the rock reaches [math: 44∘C] at 190 m depth (assume [math: 0∘C] at the surface). The thermal conductivity of the intervening rock is 4.75[math: Wm−1∘C−1] and the latent heat of fusion of ice is 334 [math: kJ/kg]. a) Neglecting other heat sources, calculate the mass of ice that melts per hour due to the geothermal gradient, in [math: kg/hour] to two significant figures. You may use the formula for thermal conductivity:[math: K=QdAΔT]where [math: K] is the thermal conductivity, [math: Q] is the amount of heat transferred per unit time, [math: d] is the distance between two isothermal planes, [math: A] is the area of the surface through which the heat flows, and [math: ΔT] is the temperature difference between the two isothermal planesMelted mass of water per hour = Answer 1 Question 4[input] kg/hour [5] b) Assuming that a mass m=27kg of water is melted per hour in the pond, calculate the thickness of the melted layer at the bottom of the pond after 10 hours. Assume a density of water equal to [math: 1000kg/m3] and that the pond is cylindrical. Give the answer in millimeters to one significant figure. thickness of melted layer = Answer 2 Question 4[input] mm [5]
Question text1. A shallow circular pond of radius freezes completely during very cold nights. The pond is in a geothermal area in which the temperature of the rock reaches at 190 m depth (assume at the surface). The thermal conductivity of the intervening rock is 4.75 and the latent heat of fusion of ice is 334 . a) Neglecting other heat sources, calculate the mass of ice that melts per hour due to the geothermal gradient, in to two significant figures. You may use the formula for thermal conductivity:where is the thermal conductivity, is the amount of heat transferred per unit time, is the distance between two isothermal planes, is the area of the surface through which the heat flows, and is the temperature difference between the two isothermal planesMelted mass of water per hour = Answer 1 Question 4[input] kg/hour [5] b) Assuming that a mass m=27kg of water is melted per hour in the pond, calculate the thickness of the melted layer at the bottom of the pond after 10 hours. Assume a density of water equal to and that the pond is cylindrical. Give the answer in millimeters to one significant figure. thickness of melted layer = Answer 2 Question 4[input] mm [5]
Question text1. A shallow circular pond of radius [math: r=24m] freezes completely during very cold nights. The pond is in a geothermal area in which the temperature of the rock reaches [math: 45∘C] at 450 m depth (assume [math: 0∘C] at the surface). The thermal conductivity of the intervening rock is 4.75[math: Wm−1∘C−1] and the latent heat of fusion of ice is 334 [math: kJ/kg]. a) Neglecting other heat sources, calculate the mass of ice that melts per hour due to the geothermal gradient, in [math: kg/hour] to two significant figures. You may use the formula for thermal conductivity:[math: K=QdAΔT]where [math: K] is the thermal conductivity, [math: Q] is the amount of heat transferred per unit time, [math: d] is the distance between two isothermal planes, [math: A] is the area of the surface through which the heat flows, and [math: ΔT] is the temperature difference between the two isothermal planesMelted mass of water per hour = Answer 1 Question 2[input] kg/hour [5] b) Assuming that a mass m=36kg of water is melted per hour in the pond, calculate the thickness of the melted layer at the bottom of the pond after 7 hours. Assume a density of water equal to [math: 1000kg/m3] and that the pond is cylindrical. Give the answer in millimeters to one significant figure. thickness of melted layer = Answer 2 Question 2[input] mm [5]
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