A piece of metal at 20^oC has a mass of 60g. When it is immersed in a current of steam at 100^∘C, 0.5g of steam is condensed on it. Determine the specific heat of metal, given that the latent heat of steam = 540 cal/g.

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Solution

Let c be the specific heat of the metal.

Heat gained by the metal

Q = m c Δt

Q = 60 × c × (100 – 20)

Q = 60 × c × 80 cal

The heat released by the steam

Q = m × L

Q = 0.5 × 540 cal

By the principle of mixtures,

Heat given is equal to Heat taken

0.5 × 540 = 60 × c ×80

c = 0.056 cal/g ^∘C

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Q. The amount of heat required to convert

1 g

of ice at

- 10^{o}

C into steam at

100^{o}

C is (specific heat of ice

= 0.5

cal/g/

^{o}

C, latent heat of ice

= 80

cal/g, latent heat of steam

= 540

cal/g)

Q. Work done in converting one gram of ice at

- 10^{\circ} C

into steam at

100^{\circ} C

is -
[Given, specific heat of ice

c_{i} = 0.5 {cal g}^{- 1}^{\circ} C^{- 1}

, specific heat of water

c_{w} = 1 {cal g}^{- 1}^{\circ} C^{- 1}

, latent heat of fusion of ice

L_{f} = 80 cal/g

, & latent heat of vaporization of steam

L_{v} = 540 cal/g

]

In Joly's defferential steam calorimeter, 3 g of an ideal gas is contained in a rigid closed sphere at $20^{\circ} C .$ The sphere is heated by steam at $100^{\circ} C$ and it is found that an extra 0.095 g of steam has condensed into water as the temperature of the gas becomes constant. Calucate the specific heat capacity of the gas in J $g^{- 1}$ $K^{- 1}$ . The latent heat of vaporization of water = 540 cal $g^{- 1}$