what is the final temperature of 75 grams of water initially at 25 degrees Celcius. upon the absorption of 2800 Joules of heat? work shown or exaplained would be greatly appreciated!!!!
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so this isn't as difficult as it seems, use the formula q=mc(delta)T
they give you the heat absorbed which is 2800J and the mass of water 75 grams, and I assume they expect you to know heat capacity of water which is 4.184J/g C now inputting all that into the equation
2800J=(75g)(4.184J/g C)T
2800=313.8T divide by 313.8
8.92=T (DONT make the mistake in thinking this is the final temperature, once you have obtained this value add it to the initial temperature they provide)
25 + 8.92 = 33.92 degress C
hope that makes sense
they give you the heat absorbed which is 2800J and the mass of water 75 grams, and I assume they expect you to know heat capacity of water which is 4.184J/g C now inputting all that into the equation
2800J=(75g)(4.184J/g C)T
2800=313.8T divide by 313.8
8.92=T (DONT make the mistake in thinking this is the final temperature, once you have obtained this value add it to the initial temperature they provide)
25 + 8.92 = 33.92 degress C
hope that makes sense