Answered

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A piece of copper steel (specific heat = 490 kg K) has a mass of 300 g. If is heated to
150°C, then plunged into 4.00 kg water (specific heat = 4180 kg at 20°C, what will
be the final temperature at equilibrium?

Sagot :

samuelonum1

The Given Data were properly outline and the relation for the quantity of heated at equilibrium was used to compute the temperature as 18.84°C

Heat Capacity

Given Data

  • Mass of Copper m1 = 300g = 0.3kg
  • Specific heat of copper c1 = 490Kg K
  • Initial Temperature of Copper  T1= 150°C
  • Mass of water m2 = 4kg
  • Specific heat of water = 42180 Kg K
  • Initial Temperature of water  T1= 20°C

At equilibrium, the final temperature of the system T2 will be same for both copper and water

McΔT(copper) = McΔT(water)

0.3*490*(T2 - 150) = 4*4180*(T2-20)

147*(T2 - 150) =16720*(T2-20)

147T2 - 22050 = 16720T2 - 334400

collect like terms

147T2 -16720T2  = - 334400+22050

147T2 -16720T2  = - 334400+22050

-16573T2 = -312350

16573T2 = 312350

Divide both sides by 16573

T2 = 312350/16573

T2 = 18.84°C

Learn more about heat capacity here:

https://brainly.com/question/21406849

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