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?

Question

1203 views around the world

You can reuse this answer
Creative Commons License

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