x Course: PHYS206-01-02 Intro x 8 Thermal Energy x The Expert TAI Human-like Gr

Question

x Course: PHYS206-01-02 Intro x 8 Thermal Energy x The Expert TAI Human-like Gr x sm42sc.theexpertta.com/Common/TakeTutorialAssignment.aspx Student: ekeampbelegeoestaledu Class Management I Help (50%) Thermal Energo 2Begin Date: 27/2017 12:00:00 AM Due Date PAMEnd Date: Aa body Problem 1: A 108 kg athlete was exercising in the bot sun and became dehydrated. He stopped sweating and his average temperature rose to 1015 F Heat stroke occurs when the core body temperature is above 105 F) R In an attempt to bydrate AND cool off quickly athe same time, be quickly swallowed 13 kg of 6Fice.This is probably not a great idea. Can you imagine the brain freeze he had? What would his average temperature in Fahrenheit beafher be and the ice came to oquilibrium? Potential 4 5 6 cotano asino Lacos0 213 atan0 Lacotan0 Lsinh00 Degrees Radians deduction per food back. Hints: for a 0 deduction Hints remaining.

Explanation / Answer

Tibody = 101.5 F = 38.61   C

Tiice = 101.5 F = -14.44   C

mbody = 108kg

mice = 1.7kg

cbody = 3470J/kg*C

cice = 2050 J/kg*C

cwater = 4180 J/kg*C

Lf =3.33*10^5

By law of conservation of energy,

Heat lost by human body = heat gain by the ice

Heat lost by human body = heat required to raise the temperature of ice + heat required to phase change of ice to water + heat required to raise the temperature of water

-[mbody*cbody*T] = mice*cice*T + mice*Lf + mwater*cwater*T

-[mbody*cbody*(Tf – Tibody)] = mice*cice*(Tfmelting – Tiice) + mice*Lf + mwater*cwater*(Tf – Tiwater)

Plugging values,

-[108*3470*( Tf – 38.61)] = 1.7*2050*(0-(-14.44)) + 1.7*3.33*10^5+1.7*4180*( Tf - 0)

Calculations gives, Tf = 36.3 oC

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