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Suppose you mix 100.0 g of water at 23.0 oC with 75.0 g of water at 75.7 oC. What will be the final temperature of the mixed water, in oC? Type answer:

Respuesta :

Answer:

Explanation:

Let final temperature be T.

Heat lost by hot water

mass x specific heat x fall in temperature

75 x s x ( 75.7 - T )

Heat gained by cool water

= 100 x s x ( T - 23 )

heat lost= heat gained

75 x s x ( 75.7 - T )   = 100 x s x ( T - 23 )

5677.5 - 75 T= 100 T- 2300

175 T= 7977.5

T = 45.6⁰C .

onyebuchinnaji

The final temperature of the mixed water is 45.6 ⁰C.

The given parameters;

  • mass of the cold water, m = 100 g
  • initial temperature of the water, t₁ = 23 ⁰C
  • initial  temperature of the hot water, t₂ = 75.7⁰ C
  • mass of the hot water = 75 g
  • specific heat capacity of water is 4.184 J/g⁰C

The final temperature of the mixture is calculated as follows;

Based on the principle of conservation of energy;

Heat lost by hot water = heat gained by the cold water

mcΔθ₂ = mcΔθ₁

75 x 4.184 x (75.7 - T) = 100 x 4.184 x (T - 23)

75 x (75.7 - T) = 100 x (T - 23)

(75.7 - T) = 1.333(T - 23)

75.7 - T = 1.333T - 30.659

75.7 + 30.659 = 1.333T + T

106.359  = 2.333T

[tex]T = \frac{106.359}{2.333} \\\\T = 45.6 \ ^0C[/tex]

Thus, the final temperature of the mixed water is 45.6 ⁰C.

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