Answer to Question #166237 in General Chemistry for Danielle

Question #166237

how many grams of water were present if 6.5 KJ of heat energy raise the temperature from 34-56°C specific heat of water equals 4200 J/Gtimes Celsius


1
Expert's answer
2021-02-24T03:41:38-0500

Here the formula we will use =

"\\Rightarrow" Q=mc"(T_2-T_1)"


where Q= Heat energy raised

m=mass of water

c=specific heat

"T_1" =initial temperature

"T_2" = final temperature


Now according to question

Q=6.5KJ=6500J

"T_1" =34 degree celsius

"T_2" =56 degree celsius

c=4200J/Gtimes Celsius


So from the above mention formulla

mass of water (m)= "\\dfrac{Q}{c\\times(T_2-T_1)}"


m="\\dfrac{6500}{4200\\times(56-34)}"


m=0.07 gm of water.


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