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In MLT θ system (T being time and θ temperature), what is the dimension of thermal conductivity?
- a)ML−1T −1θ −3
- b)MLT −1θ −1
- c)MLθ −1T−3
- d)MLθ −1T−2
Correct answer is option 'C'. Can you explain this answer?
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In MLT θ system (T being time and θ temperature), what is th...
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In MLT θ system (T being time and θ temperature), what is th...
Thermal conductivity is a property that describes the ability of a material to conduct heat. It is defined as the amount of heat energy transferred through a unit area of a material per unit time, per unit temperature gradient. The MLT system is a system of units where M represents mass, L represents length, and T represents time.
The dimension of thermal conductivity can be determined by analyzing the equation that defines it. The equation for thermal conductivity is:
k = Q / (A * ΔT / Δx)
where:
- k is the thermal conductivity
- Q is the heat energy transferred
- A is the cross-sectional area of the material
- ΔT is the temperature difference across the material
- Δx is the thickness of the material
Breaking down the equation, we can determine the dimensions of each term:
- The dimension of heat energy transferred (Q) is ML2T-2. This can be determined by analyzing the equation for heat energy (Q = mcΔT), where m is mass, c is specific heat capacity, and ΔT is temperature difference. The dimension of specific heat capacity is L2T-2Θ-1.
- The dimension of cross-sectional area (A) is L2.
- The dimension of temperature difference (ΔT) is Θ.
- The dimension of thickness (Δx) is L.
By substituting the dimensions into the equation for thermal conductivity, we get:
k = (ML2T-2) / (L2 * Θ / L)
Simplifying the equation, we get:
k = ML1T-3
Therefore, the dimension of thermal conductivity in the MLT system is ML1T-3, which corresponds to option 'C' in the given choices.
The dimension of thermal conductivity can be determined by analyzing the equation that defines it. The equation for thermal conductivity is:
k = Q / (A * ΔT / Δx)
where:
- k is the thermal conductivity
- Q is the heat energy transferred
- A is the cross-sectional area of the material
- ΔT is the temperature difference across the material
- Δx is the thickness of the material
Breaking down the equation, we can determine the dimensions of each term:
- The dimension of heat energy transferred (Q) is ML2T-2. This can be determined by analyzing the equation for heat energy (Q = mcΔT), where m is mass, c is specific heat capacity, and ΔT is temperature difference. The dimension of specific heat capacity is L2T-2Θ-1.
- The dimension of cross-sectional area (A) is L2.
- The dimension of temperature difference (ΔT) is Θ.
- The dimension of thickness (Δx) is L.
By substituting the dimensions into the equation for thermal conductivity, we get:
k = (ML2T-2) / (L2 * Θ / L)
Simplifying the equation, we get:
k = ML1T-3
Therefore, the dimension of thermal conductivity in the MLT system is ML1T-3, which corresponds to option 'C' in the given choices.
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In MLT θ system (T being time and θ temperature), what is th...
Unit of thermal conductivity is W/m*K watt is equal to joule per second joule is equal to Newton into metre so dimension of joule is M1L2T-2. SO DIMENSION OF WATT IS EQUAL TO M1L2T-3. THERMAL CONDUCTIVITY = M1L2T-3/((L1)*(theta1)) =. M1L1 K-1 T-3
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In MLT θ system (T being time and θ temperature), what is the dimension of thermal conductivity?a)ML−1T −1θ −3b)MLT −1θ −1c)MLθ −1T−3d)MLθ −1T−2Correct answer is option 'C'. Can you explain this answer?
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