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Find the initial temperature of 45.0g of water if its final Temperature after submerging a metal with the mass of 8.5g is 22.0°C . The initial temperature of metal before submerging it to water is 82.0°C

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onyebuchinnaji

The initial temperature of the water that resulted in the final temperature of the water-metal mixture is 20.7 ⁰C.

"Your question is not complete, it seems to be missing the following information;"

the specific heat capacity of the metal is 0.45 J/g⁰C.

The given parameters;

  • mass of water, [tex]m_w[/tex] = 45 g
  • final temperature of the water, [tex]t_w[/tex] = 22 ⁰C
  • mass of the metal, m = 8.5 g
  • initial temperature of the metal, t = 82 ⁰C.
  • specific heat capacity of the metal, c = 0.45 J/g⁰C.

The initial temperature of the water will be calculated by applying the principle of conservation of energy;

heat gained by water = heat lost by metal

[tex]Q_{water} = Q_{metal}[/tex]

[tex]m_wc_w \Delta t_w = mc\Delta t[/tex]

where;

[tex]c_w[/tex] is the specific heat capacity of the water = 4.184 J/g⁰C.

Substitute the given values;

45 x 4.184 x (22 - t) = 8.5 x 0.45 x (85 - 22)

4142.16  - 188.28t = 240.98

188.28t  = 4142.16  - 240.98

188.28t  = 3901.18

[tex]t = \frac{3901.18}{188.28} \\\\t = 20.7 \ ^0 C[/tex]

Thus, the initial temperature of the water that resulted in the final temperature of the water-metal mixture is 20.7 ⁰C.

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