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Physics

Calorimeter

A heating coi...

Question

A heating coil is immersed in a calorimeter of heat capacity $50 J C^{- 1}$ containing $1.0$ kg of a liquid of specific heat capacity $450 J K g^{- 1} C^{- 1}$ . The temperature of liquid rises by $10^{o}$ C when $2.0$ A current is passed for the $10$ minutes. Find the resistance of the coil.

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Solution

$V \times 2 \times 10 \times 60 = (450 \times 10) + 50 \Rightarrow V = \frac{4550}{1200} = 3.791 V R = \frac{V}{I} = \frac{3.791}{2} 1.895 Ω$

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A heating coil is immersed in a calorimeter of heat capacity 50JC−1 containing 1.0 kg of a liquid of specific heat capacity 450JKg−1C−1. The temperature of liquid rises by 10oC when 2.0 A current is passed for the 10 minutes. Find the resistance of the coil.

V×2×10×60=(450×10)+50⇒V=45501200=3.791VR=VI=3.79121.895Ω

A heating coil is immersed in a calorimeter of heat capacity $50 J C^{- 1}$ containing $1.0$ kg of a liquid of specific heat capacity $450 J K g^{- 1} C^{- 1}$ . The temperature of liquid rises by $10^{o}$ C when $2.0$ A current is passed for the $10$ minutes. Find the resistance of the coil.

$V \times 2 \times 10 \times 60 = (450 \times 10) + 50 \Rightarrow V = \frac{4550}{1200} = 3.791 V R = \frac{V}{I} = \frac{3.791}{2} 1.895 Ω$