A physicist needs to measure room temperature. Not having anything better at han

Question

A physicist needs to measure room temperature. Not having anything better at hand, she decides to make use of the thermal expansion of water to construct a thermometer.She calibrates her apparatus to a dependable thermometer at 10 deg C and 40 deg C, and divides the range between these temperatures into 30 equal intervals.what will be the true temperature when her equipment reads 25 deg C ? she knows that the volume of water can be described in this temperature range by the formula

V= V0(1- 2.525 * 10-4 +4.98*10-6 T+ 4.94*10-6T2) Where T is the temperature in oC.

Explanation / Answer

please rate: the volume of water at 10 degrees = V10 = Vo(1 - 2.525*10^4 + 4.98 * 10^5 + 4.94* 10^4) = 1.0002913 Vo. the volume of water at 40 degrees = V40 = Vo(1 - 2.525*10^4 + 4*4.98 * 10^5 + 16*4.94* 10^4) =1.0078507 Vo. each division of thermometer corresponds to (40-10)/30 = 1 degree C. for 25 degrees , number of division of the water level = 25-10 = 15. the volume of water in the thermometer is V25 = V10 + (V40 - V10)*(15/30) = (V40+V10)/2 = 1.004071 Vo let the true temperature be T. then , V = 1.004071 Vo= Vo(1- 2.525 * 10-4 +4.98*10-6 T+ 4.94*10-6T2) =>4.94*10-6T2 + 4.98*10-6 T - 0.0043235 = 0. => T = 29.084 , -30.0921 hence true temperature is : 29.084 degrees C.

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