Physics.... session masteringphysics.com Spring 2015, PHYS 1111 1113, Sections A

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Physics....

session masteringphysics.com Spring 2015, PHYS 1111 1113, Sections A/BB-EE signed in as Brandon Cole Help HW 13 Ch-14 Problem 14,46 previous 11 of 14 Problem 14.46 Part A Alaboratory technician drops an 88.5g solid Compute the specific heat capacity of the sample. sample of unknown material at a temperature of 100 (Assume no heat loss fo the surroundings.) C into a calorimeter. The calorimeter can is made of 0.155kg of copper and contains 0.200kg of water, and both the can and water are initially at 21.0 C. The final temperature of the system is measured to be 25.9 J (kg K) Submit My Answers up Incorrect: Try Again: 5 attempts remaining MacBook Air

Explanation / Answer

apply the principle of calorimetrry as

HEat lost by solid = Heat gained by Calorimeter + heat gained by water

change of temp o solid is (100 - 25.9) = 74.1 deg C

change of temp of water = 25.9 - 21   = 4.9 deg C

change of temp of calorimeter = 25.9 -21 = 4.9 deg C

sp heat of calorimert copper = 387 J/.kg C

sp heat of water = 4186 J/kg C

so

so MCDT of solid = mcDT of water + mcDT of calorimeter

88.5 * c * 74.1 = (0.155 * 387 * 4.9) + (0.2* 4186 * 4.9)

c = sp heat of solid = 4396.2065/(88.5 * 74.1)

c = 0.67 J/kg C--------------------<<<<<<<<Answer

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