How many pellets must be added to 450 g of water that is initially at 20.0°C to make the equilibrium temperatu

Question

Lead pellets, each of mass 2.00 g, are heated to 200°C. How many pellets must be added to 450 g of water that is initially at 20.0°C to make the equilibrium temperature 25.0°C? Neglect any energy transfer to or from the container.

Answer 1

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Suppose number of Lead pellets added =N

Mass of each Lead pellets = m = 2.00 g

Mass of N Lead pellet = m *N= 2.00*N g

Initial temperature of Lead pellets = tp1= 200.0°C

Final temperature of pellets = tp1= 25.0°C

Fall in temperature of pellets =tp1 - tp2=200.0 - 25.0=175.0°C

Specific heat capacity of Lead = Sl = 0.127 J/g/C

Heat lost by N pellets = H=m*N*Sl*(tp1 - tp2)

Heat lost by N pellets = H==2.00*N*0.127*175.0

Heat lost by N pellets = H =44.45*N J................(1)

Mass of water =Mw =450 g

Initial temperature of water = t1= 20.0°C

Final temperature of water = t2= 25.0°C

Rise in temperature of water =t2 - t1 = 25.0 - 20.0 = 5.0 °C

Specific heat of water = S= 4.181J/g/C

Heat gained by water =Mw*S*(t2-t1)

Heat gained by water =450*4.181*5

Heat gained by water =9407.25 J.....................(2)

From principle of heat exchange,

Heat lost by Lead pellets =Heat gained by water

44.45*N =9407.25

N =9407.25 / 44.45

N =212

212 Lead pellets must be added to 450 g of water that is initially at 20.0°C to make the equilibrium temperature 25.0°C
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Answer 2

Well, you're going to need to find the specific heat of lead.

Start by calculating the amount of heat you'll need to add to the water to raise its temperature from 20 to 25 C.

q=mcDT = 450 X 4.184 X 5 = 9414 Joules

Once you have the specific heat of lead, use the same equation with the values for lead to determine how many Joules of heat each pellet will provide:

q = 2.00 X _____ J/gC X 175 C = _____ J/pellet

Finally, divide 9414 by the value you just calculated to give you the number of pellets of lead you will need.

Answer 3

heat energy = mass* specific heat * temp difference.

Heat required raising the temperature of water
from 20° to 25° is
450*1* [25 -20] = 2250 cal.

Heat lost by 2N g of lead is 2N*[0.03] [200 - 25]
= 10.5 N where N is the number of pellets added.

Heat lost = heat gained.
10.5 N = 2250
N = 214 pellets.

Answer 4

q = cmdT qgain = qlost c1mdT (water) = c2mdT (lead) dT (water) = 25 - 20 dT (lead) = 200 - 25 m (lead) = 490 g m (water) = unknown c1 = water c2 = lead c for water and lead are constants, but I don't remember them, so you look them up. m(water) = (c1/c2) * (dT(lead)/dT(water)) * m(lead) m(water) / 0.80 g = # pellets