The initial and final temperatures of water as recorded by an observer are 4-06 - 0-2oC and 78-9 - 0-3oC- Calculate the rise in temperature with proper error limits-

θ_1 = (40.6 ±0.2)^oC, θ_2 = (78.3 ± 0.3)^oC Rise in temperature, θ = θ_2 θ_1 = 78.3 40.6 = 37.7^oCθ = ±(θ + θ_2) = ±(0.2 + 0.3) = 0.5^oC Hence, ri ...

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Standard XII

Physics

Error and Uncertainty

The initial a...

Question

The initial and final temperatures of water as recorded by an observer are $(4.06 \pm 0.2)^{o} C$ and $(78.9 \pm 0.3)^{o} C$ . Calculate the rise in temperature with proper error limits.

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Solution

$θ_{1} = (40.6 \pm 0.2)^{o} C, θ_{2} = (78.3 \pm 0.3)^{o} C$
Rise in temperature, $θ = θ_{2} - θ_{1} = 78.3 - 40.6 = {37.7}^{o} C △ θ = \pm (△ θ + △ θ_{2}) = \pm (0.2 + 0.3) = {0.5}^{o} C$
Hence, rise in temperature = $(37.7 \pm 0.5)^{o} C$

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The initial and final temperatures of water as recorded by an observer are (4.06±0.2)oC and (78.9±0.3)oC. Calculate the rise in temperature with proper error limits.

θ1=(40.6±0.2)oC,θ2=(78.3±0.3)oCRise in temperature, θ=θ2−θ1=78.3−40.6=37.7oC△θ=±(△θ+△θ2)=±(0.2+0.3)=0.5oCHence, rise in temperature = (37.7±0.5)oC

The initial and final temperatures of water as recorded by an observer are $(4.06 \pm 0.2)^{o} C$ and $(78.9 \pm 0.3)^{o} C$ . Calculate the rise in temperature with proper error limits.

$θ_{1} = (40.6 \pm 0.2)^{o} C, θ_{2} = (78.3 \pm 0.3)^{o} C$
Rise in temperature, $θ = θ_{2} - θ_{1} = 78.3 - 40.6 = {37.7}^{o} C △ θ = \pm (△ θ + △ θ_{2}) = \pm (0.2 + 0.3) = {0.5}^{o} C$
Hence, rise in temperature = $(37.7 \pm 0.5)^{o} C$