Question

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A copper sphere of mass 500 g is heated to 100 °C and then introduced into a copper calorimeter containing 100 g of water at 20 °C. Find the maximum temperature of the mixture, if the mass of the calorimeter is 100 g and the specific heat of the calorimeter is 0.1 cal/(g.°C).


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Answer 1

$<b>• SOLUTION •$

• Data : m = 500 g, c = 0.1 cal/(g.°C), T′ = 100 °C, m¹ = 100 g, c¹ = 1 cal/(g.°C), T¹ = 20 °C, m² = 100 g, c² = 0.1 cal/(g. °C), T² = 20 °C, T = ?

Heat lost by the sphere = heat gained by the water and the calorimeter.

· : mc (T′ - T) = m¹c¹ (T - T¹) + m²c² (T - T²)

= 500 g × 0.1 cal/(g. °C) × (100 °C - T)

= 100 g × 1 cal/(g. °C) × (T - 20 °C) +
100 g × 0.1 cal/(g. °C) × (T - 20 °C)

= 50 (100 °C - T) = 110 × (T - 20 °C)

= 500 °C - 5T = 11T - 220 °C

= 16T = 720 °C

T =
$\frac{720 \: }{16} \: = \: 45$

Maximum temperature of the mixture = 45 °C

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Answer 2

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