How much heat is required to convert $10 g$ $10g$ of ice at $- 10^{\circ} C$ $-10^@C$ into steam at $100^{\circ} C$ $100^@C$ ?

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Answer 1 · 2017-04-10T16:47:25

$7300$ $7300$ calories

When $10 g$ $10g$ of ice at $-20°C$ s being converted into steam at $100°C$ , there are four stages.

Ice at $-20°C$ to ice at $0°C$ - here it continues to be in the same state i.e. ice and hence heat required is $"mass"xx"specific heat"xx"change in temperature"$ Specific heat for ice is $0.5$ cal/g-°C.
Then from ice at $0°C$ to water at $0°C$ and heat required is $"mass"xx"latent heat"$ , this latent heat required is for conversion of each unit mass of substance. From ice to water it is $80"cal per gram"$
Water at $0°C$ to water at $100°C$ - here it continues to be in the same state i.e. water and hence heat required is $"mass"xx"specific heat"xx"change in temperature"$ Specific heat for water is $1$ cal/g-°C.
Then from water at $100°C$ to steam at $100°C$ and heat required is $"mass"xx"latent heat"$ , this latent heat required is for conversion of each unit mass of substance. From water to steam, it is $540"cal per gram"$

Hence, to find heat is required to convert $10g$ of ice at $-20°C$ into steam at $100°C$ ,

first calculate heat required to convert $10g$ of ice at $-20°C$ to ice at $0°C$ . As specific heat of ice is $0.5"cal/g°C"$ , this is

$10xx20xx0.5=100$ cal.

heat required to convert $10g$ of ice to $10g$ of water at $0°C$ is

$10xx80=800$ cal - as latent heat is $80"cal/g"$

heat required to convert $10g$ of water at $0°C$ to $10g$ of water at $100°C$ is

$10xx100x1=1000$ cal

heat required to convert $10g$ of water at $1000°C$ to $10g$ of steam at $100°C$ is

$10xx540=5400$ cal - as latent heat is $540"cal/g"$

Hence, total heat required is $100+800+1000+5400=7300$ calories

How much heat is required to convert 10g10g10g of ice at -10^@C−10∘C-10^@C into steam at 100^@C100∘C100^@C?