WILL REWARD MEDAL/FAN!!!((: A blacksmith heat an iron bar to 1445 Celsius. The blacksmith then tempered the metal by dropping into 42,800 cm^3 of water that had a temperature of 22Celsius. The final temperature of the system was 45 Celsius. What was the mass of the bar?

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WILL REWARD MEDAL/FAN!!!((: A blacksmith heat an iron bar to 1445 Celsius. The blacksmith then tempered the metal by dropping into 42,800 cm^3 of water that had a temperature of 22Celsius. The final temperature of the system was 45 Celsius. What was the mass of the bar?

Mathematics
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so the temperature of the water has risen by \(45-22 = 23^{\circ}C\)
that's the delta t right?
Yes, then the heat added to the water is given by, \(Q=c*m*\Delta t=4.18*42800*23 = 4114792 J\)

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That means the bar has lost the same amount of heat. You can use that to find the mass of bar
Oh I see.
so basically it will be 4114792 = mass x 0.446 x (1445-45) Celsius which gives you the mass is 65890 grams or 6.6 kilograms?
0.446 is specific heat of iron is it ?
yeah
Looks good !
thanks can you also help me with this one?
a 752 cm^3 sample of water was placed in a 1 kg aluminum pan. The initial temperature of the pan was 26 Celsius, and the final temperature of the system was 39 Celsius. What was the initial temperature of the water?
looks like a similar problem.. give it a try
well i got stuck because i wasn't sure 0.902 j/ 1gC x (39- delta t) x 1000g = 752 cm^3 totally got that wrong i'm guessing
First, write down the given info Water : \(m = 752 cm^3 = 752 g\) \(c = 4.18 J/G^{\circ}C\) \(\Delta t = ?\) Aluminium pan : \(m = 1kg= 1000 g\) \(c = 0.9 J/G^{\circ}C\) \(\Delta t = 13\)
how do we know what the delta t is if we don't have the q of the equation ( i think that is the energy released)?
Energy released by water = Energy absorbed by aluminium pan
\[4.18*752*\Delta t = 0.9*1000*13 \] you can solve \(\Delta t\)
oh okay and would that be 39-26?
wait hold on
oh never mind so now you just basically solve for the delta t right and the answer would be in celsius?
what do you get for \(\Delta t\) ?
3.72 Celsius...
good \(\Delta t = 3.72 \) and you know that final temperature = \(39\) so initial temperature = ?
so subtract?
careful, water has lost some heat that means the initial temperature is greater than the final temperature
oh oops i read it wrong i mean add because the final temperature was the result of the two temperatures i accidentally interpreted it wrong 42.72 Celsius?
Yep, 42.72 C looks good
The fuel value of peanuts is 25 kj/ gram. If an average adult needs 2800 kilocalories of energy a day, what mass of peanuts would meet an average adult's energy needs for the day? Assume all of the fuel value of the peanuts can be converted to useful energy.
I am confused about what calories are.
dont get confused, all the problems you have been working on are similar... if we don't know what calorie means, we can always google :)
the main reason i suck at these problems is that my teacher just gives it to me along with the specific heat formula, so I don't get almost everything
Yes, that is all we ever need to work these problems First things first, lets put down the given info
25 kj/ grams 2800 kilocalories of energy a day we are finding mass
google gave me 1 kilocalorie = 4184 joules
yeah i think a calorie is like 4.18 something per joules.
right, so 2800 kilocalories = how many kilojoules ?

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