Calculate the heat required to convert nbsp- 3 kg of ice at - 12 - C kept in a calorimeter to steam at 100 - at atmospheric pressure- Given - specific heat of ice - 2100 - 10 5 J kg - 1 K - 1 - specific heat of water - 4186 - 10 3 J kg - 1 K - 1 - latent heat of fusion of ce - 3-35 - 10 5 J kg - 1 and latent heat of steam - 2-256 - 10 6 J kg - 1-

Step 1: Bringing ice at 0 #176; C #160;Q1 = MC #916;T= gt;3*2100*(12 0)= gt; 75600KgJStep 2: Changin to water at 0 #176;CQ2=ML= gt;3*3.35*10= gt ...

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Byju's Answer

Standard VIII

Physics

Evaporation

Calculate the...

Question

Calculate the heat required to convert  $3 k g$ of ice $a t - 12^{\circ} C$ kept in a calorimeter to steam at $100^{\circ}$ at atmospheric pressure. Given : specific heat of ice $= 2100 \times 10^{5} J {k g}^{- 1} K^{- 1}$ , specific heat of water $= 4186 \times 10^{3} J {k g}^{- 1} K^{- 1}$ , latent heat of fusion of $c e = 3.35 \times 10^{5} J {k g}^{- 1}$ and latent heat of steam $= 2.256 \times 10^{6} J {k g}^{- 1}$ .

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Solution

Step 1: Bringing ice at 0° C
Q1 = MCΔT
=>32100(12-0)
=> 75600KgJ
Step 2: Changin to water at 0°C
Q2=ML
=>33.3510
=>100.5Jkg
Step 3: Bringing water to 100°C
Q3=MCΔT
=>34.186100
=>1255.8JKg
Step4: Converting water to steam
Q4=ML
=>3*22.576JKg
=>67.728JKg
Total heat == Q1+Q2+Q3+Q4
==> 77024.03J

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Similar questions

$Calculate the amount of heat required to convert 1.00 kg of ice at -10°C$
$at normal pressure to steam at 100°C.$

$The specific heat capacity of ice = 2100 J k g^{- 1} K^{- 1}$

$Latent heat of fusion of ice = 3.36 \times 10^{5} J k g^{- 1}$

$The specific heat capacity of water = 4200 J k g^{- 1} K^{- 1}$

$Latent heat of vaporization of water = 2.25 \times 10^{6} J k g - 1$

Q. A copper calorimeter of mass

100 g m

contains

200 g m

of a mixture of ice and water. Steam at

100^{\circ} C

under normal pressure is passed into the calorimeter and the temperature of the mixture is allowed to rise to

50^{\circ} C

. If the mass of the calorimeter and its contents is now

330 g m

, what was the ratio of ice and water in the beginning? Neglect heat losses.
Given: Specific heat capacity of copper

= 0.42 \times 10^{3} J k g^{- 1} K^{- 1}

,
Specfic heat capacity of water

= 4.2 \times 10^{3} J k g^{- 1} K^{- 1}

Specific heat of fusion of ice

= 3.36 \times 10^{5} J k g^{- 1}

Latent heat of condensation of steam

= 22.5 \times 10^{5} J k g^{- 1}

Q. If

10 g

of ice is added to

40 g

of water at

15^{o} C

, then the temperature of the mixture is (specific heat of water

= 4.2 \times 10^{3} J k g^{- 1} K^{- 1}

,Latent heat of fusion of ice

= 3.36 \times 10^{5} J k g^{- 1}

)

Q. Calculate the total amount of heat energy required to convert

100 g

of ice at

- 10^{\circ} C

completely into water at

100^{\circ} C

. Specific heat capacity of ice =

2.1 J g^{- 1} K^{- 1}

, specific heat capacity of water =

4.2 J g^{- 1} K^{- 1}

, specific latent heat of ice =

336 J g^{- 1}

Q. The amount of heat required to convert 1 g of ice (specific 0.5 cal at

g^{- 1 o} C^{- 1}

) at

- 10^{0} C

to steam at

100

^{\circ} C

is ___________.

[ Given: Latent heat of ice is

80 C a l / g m,

Latent heat of steam is

540 C a l / g m

, Specific heat of water is

1 C a l / g m / C

]

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Step 1: Bringing ice at 0° C Q1 = MCΔT=>3*2100*(12-0)=> 75600KgJStep 2: Changin to water at 0°CQ2=ML=>3*3.35*10=>100.5JkgStep 3: Bringing water to 100°CQ3=MCΔT=>3*4.186*100=>1255.8JKgStep4: Converting water to steamQ4=ML=>3*22.576JKg=>67.728JKgTotal heat == Q1+Q2+Q3+Q4==> 77024.03J

Step 1: Bringing ice at 0° C
Q1 = MCΔT
=>32100(12-0)
=> 75600KgJ
Step 2: Changin to water at 0°C
Q2=ML
=>33.3510
=>100.5Jkg
Step 3: Bringing water to 100°C
Q3=MCΔT
=>34.186100
=>1255.8JKg
Step4: Converting water to steam
Q4=ML
=>3*22.576JKg
=>67.728JKg
Total heat == Q1+Q2+Q3+Q4
==> 77024.03J