An object with a mass of 150 g is dropped into 500 mL of water at 0^@C. If the object cools by 72 ^@C and the water warms by 3 ^@C, what is the specific heat of the material that the object is made of?

1 Answer
Oct 14, 2017

Probably Scandium (Sc) at 0.58111J/(g°C)

Explanation:

Q=mc∆T
=>c=Q/(m∆T)

For water:

What we know-

  • c=4.184J/g°C
  • m=500g
  • ∆T=3°C

therefore Q=4.184J/(g°C)*500g*3°C

=>Q=6276J

For object:

What we know-

  • m=150g
  • ∆T=-72°C (The object cools)

Since heat gained (Q) by water is the heat lost (-Q) by the object,

Q_"water"=-Q_"object"

So, c=(-6276J)/(150g*-72°C)

=>c=0.58111J/(g°C)

Which is a little over the the specific heat of Scandium (Sc) which is 0.5677J/(g°C).

But as we have made a lot of assumptions over here, the material is probably Scandium.