CHEMISTRY HIGH SCHOOL

A 134.0 g sample of an unknown metal is heated to 91.0⁰C and then placed in 125 g (125 mL) of water at 25.0⁰C. The final temperature of the water is measured at 31.0⁰C. Calculate the specific heat capacity of the unknown metal.

Answers

Answer 1

Answer:

Answer:

The specific heat capacity of the metal is approximately 0.3903 J/(g·°C)

Explanation:

The mass of the sample of the unknown metal, $m_m$ = 134.0 g

The temperature to which the metal is raised, $t_m$ = 91.0°C

The mass of water into which the mass of metal is placed, $m_w$ = 125 g

The temperature of the water into which the metal is placed, $t_w$ = 25.0°C

The final temperature of the water, $t_f$ = 31.0°C

The specific heat capacity of water, $c_w$ = 4.184 J/(g·°C)

The specific heat capacity of the metal = $c_m$

Therefore, by the first laws of thermodynamics we have;

The heat transferred = Heat supplied by the metal = Heat gained by the water

The heat transferred, ΔQ, is given as follows;

ΔQ = $m_w$ × $c_w$ ×( $t_f$ - $t_w$ ) = $m_m$ × $c_m$ ×( $t_m$ - $t_f$ )

125 × 4.184 × (31 - 25) = 134 × $c_m$ × (91 - 31)

∴ $c_m$ = (125 × 4.184 × (31 - 25))/(134 × (91 - 31)) ≈ 0.3903 J/(g·°C)

The specific heat capacity of the metal = $c_m$ ≈ 0.3903 J/(g·°C)

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Answers

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Answers

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Answers

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Answers

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