Im getting a negative answer for heat released = heat absorbed, does anyone know why?
im doing 20*.39*(100-T)= 50.7*4.18(T-22)
4A.9
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Justin Vayakone 1C
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Re: 4A.9
Heat released = - heat absorbed
Your work here looks right. I'm assuming you used the negative to change (T-100) into (100-T) right?
Your work here looks right. I'm assuming you used the negative to change (T-100) into (100-T) right?
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Hui Qiao Wu 1I
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Re: 4A.9
Your change in temperature for copper should be final temperature-100, not the other way around.
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Katie Kyan 2K
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Anisha Chandra 1K
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Re: 4A.9
Yeah, you can do (Tfinal - 100) and then add a negative sign to the other side with water to show that the heat lost by copper is that gained by water.
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