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A 5-kg piece of lead (specific heat 0.03 cal/gC) having a temperature of 80C is added to 500 g of water having a temperature of 20C. What is the final equilibrium temperature (in C) of the system?

Respuesta :

jcherry99

Answer:

Explanation:

Comment

The principle involved here is heat lost = heat gained

Formula

m1 * c1* delta (t1) = m2 * c2 * delta (t2)

Givens

  • m1 = 5000 grams (5 kg)
  • c1 = 0.03
  • delta t1 = (80 - t)
  • m2 = 500 g
  • c2 = 4,186
  • delta t2 = (t - 20)

Solution

5000 * 0.03 * (80 - t) = 500 * 4.186 * (t - 20)          Remove Brackets.

12000 - 150t = 2093t - 41860                                 Add 150t to both sides

12000 - 150t+150t = 3093t + 150t - 41860             Combine

12000 = 3243t - 41860                                            Add 41860 to both sides

12000 + 41860 = 3243t - 41860 +41860

53860 = 3243t                                                         Divide by 3243

53860 / 3243 = 3243t/3243

16,61 = t

This is of course not possible, but it is what the numbers give. Numbers can do anything.

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