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A resistance heater was wound around a 5...

A resistance heater was wound around a `5.0g` metallic cylinder. A current of `0.84 A` was passed through the heater for `20s` while the drop in voltage across the heater was `50V`. The temperature change of the cylinder was from `25^(@)C` before the heating period and `35^(@)C` at the end. If the heat loss is neglected, what is the specific heat of the cylinder metal in `cal g^(-1)K^(-1)`.

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To solve the problem, we need to calculate the specific heat of the metallic cylinder using the information provided. Here’s a step-by-step solution: ### Step 1: Calculate the Power Input The power input (or heat generated) by the resistance heater can be calculated using the formula: \[ \text{Power} = \text{Voltage} \times \text{Current} \] Given: ...
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