Question

M grams of steam at 100⁰C is mixed with 200g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40⁰ C[heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g ], the value of M is __.

Answer 1

Explanation:

Given, M grams of steam = 100⁰C is mixed with 200g of ice at its melting point in a thermally insulated container.

If it produces liquid water at 40⁰ C[heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g

Then, the value of M is

Here, heat absorbed by ice = ${\mathbf{m}}_{\mathbf{ice}}\mathbf{}\mathbf{L}{\mathbf{f}\mathbf{}}_{}\mathbf{+}\mathbf{}{\mathbf{m}}_{\mathbf{ice}}\mathbf{}{\mathbf{C}}_{\mathbf{w}}\mathbf{(}\mathbf{40}\mathbf{-}\mathbf{0}\mathbf{)}$

The heat released by steam =$\mathbf{}{\mathbf{m}}_{\mathbf{steam}}\mathbf{}{\mathbf{L}}_{\mathbf{v}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{m}}_{\mathbf{steam}}\mathbf{}{\mathbf{C}}_{\mathbf{w}}\mathbf{(}\mathbf{100}\mathbf{-}\mathbf{40}\mathbf{)}$

Heat absorbed = heat released

${\mathrm{m}}_{\mathrm{ice}}\mathrm{L}{\mathrm{f}}_{}+{\mathrm{m}}_{\mathrm{ice}}{\mathrm{C}}_{\mathrm{w}}(40-0)={\mathrm{m}}_{\mathrm{steam}}{\mathrm{L}}_{\mathrm{v}}+{\mathrm{m}}_{\mathrm{steam}}{\mathrm{C}}_{\mathrm{w}}(100-40)$

200×80 cal/g + 200×1 cal/g/⁰C ×(40 – 0)

= m ×540 cal/g + 540×1 cal/g/⁰C ×(100 – 40)

200[80 + 40] = m[540+ 60]

M=40gm

Therefore, M grams of steam at 100⁰C is mixed with 200g of ice at its melting point in a thermally insulated container. If it produces liquid water at 40⁰ C[heat of vaporization of water is 540 cal/g and heat of fusion of ice is 80 cal/g ], the value of M is 40gm.