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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?

- lilai3

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- ganeshie8

so the temperature of the water has risen by \(45-22 = 23^{\circ}C\)

- lilai3

that's the delta t right?

- ganeshie8

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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## More answers

- ganeshie8

That means the bar has lost the same amount of heat.
You can use that to find the mass of bar

- lilai3

Oh I see.

- lilai3

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?

- ganeshie8

0.446 is specific heat of iron is it ?

- lilai3

yeah

- ganeshie8

Looks good !

- lilai3

thanks can you also help me with this one?

- lilai3

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?

- ganeshie8

looks like a similar problem..
give it a try

- lilai3

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

- ganeshie8

First, write down the given info
Water :
\(m = 752 cm^3 = 752 g\)
\(c = 4.18 J/G^{\circ}C\)
\(\Delta t = ?\)
Aluminium pan :
\(m = 1kg= 1000 g\)
\(c = 0.9 J/G^{\circ}C\)
\(\Delta t = 13\)

- lilai3

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)?

- ganeshie8

Energy released by water = Energy absorbed by aluminium pan

- ganeshie8

\[4.18*752*\Delta t = 0.9*1000*13 \]
you can solve \(\Delta t\)

- lilai3

oh okay
and would that be 39-26?

- lilai3

wait hold on

- lilai3

oh never mind so now you just basically solve for the delta t right
and the answer would be in celsius?

- ganeshie8

what do you get for \(\Delta t\) ?

- lilai3

3.72 Celsius...

- ganeshie8

good
\(\Delta t = 3.72 \)
and you know that final temperature = \(39\)
so initial temperature = ?

- lilai3

so subtract?

- ganeshie8

careful, water has lost some heat
that means the initial temperature is greater than the final temperature

- lilai3

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?

- ganeshie8

Yep, 42.72 C looks good

- lilai3

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.

- lilai3

I am confused about what calories are.

- ganeshie8

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 :)

- lilai3

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

- ganeshie8

Yes, that is all we ever need to work these problems
First things first, lets put down the given info

- lilai3

25 kj/ grams
2800 kilocalories of energy a day
we are finding mass

- ganeshie8

google gave me
1 kilocalorie =
4184 joules

- lilai3

yeah i think a calorie is like 4.18 something per joules.

- ganeshie8

right, so 2800 kilocalories = how many kilojoules ?

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