Question

The quantity of heat required to raise the temperature of one kilogram of a metal is 2300 cal. Find the specific heat and water equivalent of the metal piece​

Answer 1

S = 1cal/g/℃

⇒ = 100cal/kg/℃

△T = 1℃

Q = ms△T

⇒ = 1×1×100×1

⇒ = 100cal

⇒ = 1Kcal

⇒ = 1 kilo-calorie.

Hence , OPTION : A (kilo-calorie).

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$S = 1cal/g/℃</p><p></p><p>⇒  = 100cal/kg/℃</p><p></p><p>△T = 1℃</p><p></p><p>Q   =   ms△T</p><p></p><p>⇒    = 1×1×100×1</p><p></p><p>⇒    = 100cal</p><p></p><p>⇒    = 1Kcal</p><p></p><p>⇒    = 1 kilo-calorie.</p><p></p><p>Hence , OPTION : A (kilo-calorie).</p><p></p><p>$

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