Question
Asked Nov 28, 2019
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A 10.2−g sample of ethylene glycol, a car radiator coolant, loses 946 J of heat. What was the initial temperature of the ethylene glycol if the final temperature is 32.5°C? (c of ethylene glycol = 2.42 J/g·K)

what is the answer in C

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Expert Answer

Step 1

Given:

Mass of ethylene glycol=10.2 g

Heat loss (q)=946 J

Final temperature=32.5oC

Specific heat capacity (c) of ethylene glycol=2.42 J/g.K

Step 2

To calculate initial temperature of ethylene glycol:

Consider an equation, where heat flow (q) of process is equal to change in enthalpy

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AH-m*c*AT q-m*c*AT where m-mass cspecific heat capacity AT=Change in temperature=Te Tefce

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Step 3

Calculating the initial temperature,

By substituting all t...

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q=m*c*AT -946J-10,2g*2,42J/g.K* ( x) (K) 946 x AT (10,2g 2.42) AT--38.32K 38.32°C since it is A

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