The initial and final temperatures of water a | vedclass

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(A) Given:Initial temperature,$	heta_1 = (40.6 \pm 0.2)^{\circ}C$Final temperature,$	heta_2 = (78.9 \pm 0.3)^{\circ}C$The rise in temperature is giv

Vedclass · Accepted Answer

$(38.3 \pm 0.5)^{\circ}C$

Vedclass · Answer

(A) Given:Initial temperature,$	heta_1 = (40.6 \pm 0.2)^{\circ}C$Final temperature,$	heta_2 = (78.9 \pm 0.3)^{\circ}C$The rise in temperature is given by $\Delta	heta = 	heta_2 - 	heta_1$.$\Delta	heta = 78.9 - 40.6 = 38.3^{\circ}C$.When subtracting two quantities,the absolute errors are added.Error in rise in temperature,$\Delta(\Delta	heta) = \Delta	heta_1 + \Delta	heta_2$.$\Delta(\Delta	heta) = 0.2 + 0.3 = 0.5^{\circ}C$.Therefore,the rise in temperature with proper error limits is $(38.3 \pm 0.5)^{\circ}C$.

The initial and final temperatures of water as recorded by an observer are $(40.6 \pm 0.2)^{\circ}C$ and $(78.9 \pm 0.3)^{\circ}C$. Calculate the rise in temperature with proper error limits.

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