An 85.0 kg track runner is overheated, and has a core temperature of 42.0 ºC. If I couldlower this athlete’s body temperature back to 37.0 ºC by using their excess heat to boilwater at 100.0 ºC, how many g of water would I need to do this?a) 555.4 gb) 786.8 gc) 1127 gd) 629.5 ge) 1745 g this question has something to do with specific heatsomeone tried doing (42.0)(85.0x10^3)(42-37) but idk pls help !!

Question
Asked Dec 7, 2019
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An 85.0 kg track runner is overheated, and has a core temperature of 42.0 ºC. If I could
lower this athlete’s body temperature back to 37.0 ºC by using their excess heat to boil
water at 100.0 ºC, how many g of water would I need to do this?


a) 555.4 g
b) 786.8 g
c) 1127 g
d) 629.5 g
e) 1745 g

 

this question has something to do with specific heat

someone tried doing (42.0)(85.0x10^3)(42-37) but idk pls help !!

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Expert Answer

Step 1

Given data:

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Mass of track runner(m, ) = 85.0 kg Initial temperature of track runner (T, ) = 42.0 °C Final temperature of track runner(T,) = 37.0°C Initial temperature of water( T ) = 25.0 °C (room temperature) %3D Final temperature of water( T,) =100.0°C

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Step 2

The heat given by track runner is equal to the heat absorbed by water.

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-4given off = labsorbed -m,Chum an bodyAT = m,CwaterAT| -тСниmаan body (7, -1,) %3 т,С авг ´human

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Step 3

The specific heat of water is 4.18 kJ/kg-oC.

The specific heat of human body is 3.5 kJ/kg-...

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- (T. – T; ) Chuman body (T; – T; ) = m.C (T, – T ) -85.0 kg -3.5 kJ/kg-°C(37.0 °C – 42.0°C) =m, ·4.18 kJ/kg-°C(100.0 °C – 25.0 °C) -т,с wa ter 1487.5 kJ = m, ·313.5 kJ/kg 1487.5 kg т, 313.5 1000 g = 4.7448 kg 1kg = 4744.8 g

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