Question

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

Answer 1

< b > • SOLUTION •<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

16

720

=45

Maximum temperature of the mixture = 45 °C

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Answer 2

Answer:

Hope the above attached file helps you