Question

5) A 70.5 gram piece of metal, initially at 255 °C, is rapidly transferred to 100.0 mL of distilled water, which is initially at 19.6 °C. The temperature of the water increases to 30.8 °C. Given that the density of the water = 1.00 g/mL, calculate the specific heat (J/g °C) of the metal. The specific heat of water is 4.184 J/g °C. 9- (SH)(m)(AT) A) 0.632 B) 3.37 C) 5.17 D) 0.0234 E) 0.296

Answer 1

Given, the density of water = 1.00g/ml.

Hence, the mass of water added = 1.00 $\times$ 100 = 100 g.

Hence, the amount of energy absorbed by the water to raise its temperature q = 4.184(J/g.oC)  $\times$ 100(g) $\times$ (30.8 - 19.6) = 4.686 KJ.

When metal is kept in water, the temperature of water increases and that of the metal decreases until they attain thermal equilibrium where the temperature of both of them are equal.

Therefore, the final temperature of the metal = 30.8 oC.

And also here, we assume no energy is lost to the surroundings, hence, energy lost by metal = energy gained by water = 4.686 KJ

therefore, C $\times$ 70.5 = (255 - 30.8) = 4686 J, where C is specific heat of the metal.

Therefore, C = 0.296 J/g.oC