Question

A piece of titanium metal with a mass of 15.7 g is heated in boiling water to 99.5oc (initial temperature of metal) and then dropped into a coffee cup calorimeter containing 75.0g of water at 21.7oc with a specific heat of 4.184 j/goc. when thermal equilibrium is reached, the final temperature is 24.3oc. calculate the specific heat of titanium.

Answer 1

The specific heat of titanium is 0.69 J/ g c.

The specific heat of titanium can be calculated by using the principle of calorimetry.

According to this principle, the heat lost by one body is equal to the heat gained by another body in contact with it.

The heat(Q) is calculated by the following formula,

Q = m(s)(ΔT)

Here m is the mass of the body

s is the specific heat of the body

ΔT is the change in temperature

Thus, the heat lost by titanium metal is equal to the heat gained by water present in the calorimeter,

(msΔT)Titanium= (msΔT)Water

Putting the given values in the above equation

15.7 x s x 99.5-24.3 = 75 x 4.184 x 24.3-21.7

s = 0.69 J/g C

Thus, the specific heat of titanium is 0.69 J/ g c.

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