FOLLOWING CONSTANTS ARE GIVEN FOR CALCULATIONS. Specific Heat of Water = 4186 J/kg °C, Specific Heat of Iron = 452 J /kg °C, Specific Heat of Ice = 2093 J/kg °C, Specific Heat of Lead = 130 J/kg °C Latent Heat of Fusion = 3.34 x 105 J/kg, Specific Heat of Copper = 385.1 J /kg.°C Density of iron = 7.874 g /cm3 Density of copper = 8.95 g /cm³ %3D 1. A cylindrical shape iron object of radius 400000 µm and length 2 m initially at 30 °C is placed in hot water at 60 °C. Calculate the heat energy received by the iron ball.

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Chapter1: Units, Trigonometry. And Vectors
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FOLLOWING CONSTANTS ARE GIVEN FOR CALCULATIONS
Specific Heat of Water = 4186 J/kg °C,
Specific Heat of Iron = 452 J /kg °C,
Specific Heat of Ice = 2093 J/kg °C,
Specific Heat of Lead = 130 J/kg °C
Latent Heat of Fusion = 3.34 x 105 J/kg,
Specific Heat of Copper = 385.1 J /kg.°C
Density of iron = 7.874 g /cm3
Density of copper = 8.95 g /cm3
1. A cylindrical shape iron object of radius 400000 µm and length
2 m initially at 30 °C is placed in hot water at 60 °C. Calculate
the heat energy received by the iron ball.
Transcribed Image Text:FOLLOWING CONSTANTS ARE GIVEN FOR CALCULATIONS Specific Heat of Water = 4186 J/kg °C, Specific Heat of Iron = 452 J /kg °C, Specific Heat of Ice = 2093 J/kg °C, Specific Heat of Lead = 130 J/kg °C Latent Heat of Fusion = 3.34 x 105 J/kg, Specific Heat of Copper = 385.1 J /kg.°C Density of iron = 7.874 g /cm3 Density of copper = 8.95 g /cm3 1. A cylindrical shape iron object of radius 400000 µm and length 2 m initially at 30 °C is placed in hot water at 60 °C. Calculate the heat energy received by the iron ball.
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