An object with a mass of $120 g$ is dropped into $640 mL$ of water at $0^@C$ . If the object cools by $36 ^@C$ and the water warms by $12 ^@C$ , what is the specific heat of the material that the object is made of?

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Answer 1 · 2018-01-23T14:32:46

1.77 Calorie gm^-1 'C^-1

To come in thermal equilibrium,heat energy released by the object is $m×s×del theta$ = $120×s×36$ Calorie.(where, $s$ is its specific heat)

And during this process,heat taken by water is $640×1×1×12$ Calorie (Here,mass of water = volume×density)

So,equating both we get, $s = 1.77$ Calorie gm^-1 'C^-1

An object with a mass of 120 g120 g is dropped into 640 mL640 mL of water at 0^@C0^@C. If the object cools by 36 ^@C36 ^@C and the water warms by 12 ^@C12 ^@C, what is the specific heat of the material that the object is made of?