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lilai3
 one year ago
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?
lilai3
 one year ago
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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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2so the temperature of the water has risen by \(4522 = 23^{\circ}C\)

lilai3
 one year ago
Best ResponseYou've already chosen the best response.1that's the delta t right?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Yes, then the heat added to the water is given by, \(Q=c*m*\Delta t=4.18*42800*23 = 4114792 J\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2That means the bar has lost the same amount of heat. You can use that to find the mass of bar

lilai3
 one year ago
Best ResponseYou've already chosen the best response.1so basically it will be 4114792 = mass x 0.446 x (144545) Celsius which gives you the mass is 65890 grams or 6.6 kilograms?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.20.446 is specific heat of iron is it ?

lilai3
 one year ago
Best ResponseYou've already chosen the best response.1thanks can you also help me with this one?

lilai3
 one year ago
Best ResponseYou've already chosen the best response.1a 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
 one year ago
Best ResponseYou've already chosen the best response.2looks like a similar problem.. give it a try

lilai3
 one year ago
Best ResponseYou've already chosen the best response.1well 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
 one year ago
Best ResponseYou've already chosen the best response.2First, 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
 one year ago
Best ResponseYou've already chosen the best response.1how 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
 one year ago
Best ResponseYou've already chosen the best response.2Energy released by water = Energy absorbed by aluminium pan

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2\[4.18*752*\Delta t = 0.9*1000*13 \] you can solve \(\Delta t\)

lilai3
 one year ago
Best ResponseYou've already chosen the best response.1oh okay and would that be 3926?

lilai3
 one year ago
Best ResponseYou've already chosen the best response.1oh never mind so now you just basically solve for the delta t right and the answer would be in celsius?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2what do you get for \(\Delta t\) ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2good \(\Delta t = 3.72 \) and you know that final temperature = \(39\) so initial temperature = ?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2careful, water has lost some heat that means the initial temperature is greater than the final temperature

lilai3
 one year ago
Best ResponseYou've already chosen the best response.1oh 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
 one year ago
Best ResponseYou've already chosen the best response.2Yep, 42.72 C looks good

lilai3
 one year ago
Best ResponseYou've already chosen the best response.1The 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
 one year ago
Best ResponseYou've already chosen the best response.1I am confused about what calories are.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2dont 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
 one year ago
Best ResponseYou've already chosen the best response.1the 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
 one year ago
Best ResponseYou've already chosen the best response.2Yes, that is all we ever need to work these problems First things first, lets put down the given info

lilai3
 one year ago
Best ResponseYou've already chosen the best response.125 kj/ grams 2800 kilocalories of energy a day we are finding mass

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2google gave me 1 kilocalorie = 4184 joules

lilai3
 one year ago
Best ResponseYou've already chosen the best response.1yeah i think a calorie is like 4.18 something per joules.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2right, so 2800 kilocalories = how many kilojoules ?
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