When a 24.0 g of a metal sample is heated in water at 100.0degreeC, then transfo

Question

When a 24.0 g of a metal sample is heated in water at 100.0degreeC, then transformed into 50.0 g sample of liquid water at 25.0 C in a coffee-cup calorimeter, the final temperature of the water/metal was 31.9 0degreeC. The specific heat capacity of the metal is J/g-K. The specific heat of liquid water is, 4.18 J/g-K. Select one a. 1.74 b. 0.885 c. 0.060 d. 2.78 e. 1.34 A 50.0-g sample of liquid water at 25.0 C is mixed with 29.0 g of water at45.0 degreeC. The final temperature of the water is degreeC. The specific heat capacity of liquid water is Remember heat gained = - heat lost Select one: a. 102 b. 32.3 c. 142 d. 35.0 e. 27.6

Explanation / Answer

a) heat lost by metal = heat gained by water

Q = mcdT

specific heta of water = 4.186 J/g/oC

t = final temperature = 31.9 oC

   mc(t1-t) = mc(t-t2)

   24 x c x (100-31.9) = 50 x 4.186 x (31.9-25)

On simplification,

    c = 0.885 J/g/oC = 0.885 J/g/K

Therefore,

specific heat of metal = 0.885 J/g/K

b)

heat lost by water at 45oC= heat gained by water at 25oC

Q = mcdT

c = specific heta of water

t = final temperature

   mc(t1-t) = mc(t-t2)

    29 x c x (45-t) = 50 x c x (t-25)

1305 - 29t = 50t-1250

   79t = 2555

   t = 32.3 oC

Therefore,

final temperature of the water = 32.3 oC

   

   24 x c x (100-31.9) = 50 x 4.186 x (31.9-25)

On simplification,

    c = 0.885 J/g/oC = 0.885 J/g/K

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