Question

2. 2075 Set B Q.No. 10a A mixture of 500g water and 100 g ice at
0°C is kept in a copper calorimeter of mass 200 g. How
much steam from the boiler be passed to the mixture so that
the temperature of the mixture reaches 40°C?
[4]​

Answer 1

Given info : A mixture of 500g water and 100 g ice at 0°C is kept in a copper calorimeter of mass 200 g.

To find : amount of steam from the boiler be passed to the mixture so that the temperature of the mixture reaches 40°C.

solution : specific heat of copper = 0.40 J/g.°C

specific heat of water = 4.2J/g.°C

heat of fusion of ice = 336 J/g

heat of vaporisation = 2257 J/g

case 1 : heat required to melt the ice and then increase its temperature upto 40° , H₁ = mLf + ms∆T

= 100g × 336 J/g + 100g × 4.2 J/g°C × 40°C

= 33600 + 420 × 40

= 33600 + 16800

= 50400 J

= 50.4 kJ

case 2 : heat required to increase the temperature of water , H₂ = Ms∆T

= 500g × 4.2J/g°C × 40°C

= 2100 × 40 J

= 84 kJ

case 3 : amount of heat required to increase the temperature of copper, H₃ = 200g × 0.4 J/g°C × 40

= 80 × 40 J

= 3200 J

= 3.2 kJ

now total amount of heat required = H₁ + H₂ + H₃

= 50.4 + 84 + 3.2

= 137.6 kJ

let M' mass of steam is added to the system of water , ice and copper.

so, heat released by steam = M'Lv = M' × 2257 J

from calorimetry,

heat lose = heat gain

⇒M' × 2257 = 137.6 kJ

⇒M' = 137.6 × 1000/2257 ≈ 61 g

therefore 61 g of steam should be passed to the mixture so that the temperature of the mixture reaches 40°C.

Answer 2

Given:

Given that, a combination of 500 g of water and 100 g of ice at 0°C is deposited in a copper calorimeter with a mass of 200 g.

To find:

We ought to figure out how much steam is added to the mixture from the boiler so that the mixture's temperature exceeds 40°C.

Solution:

The Old Thermodynamics says:

We know that the water, ice and copper measuring system square measure at 0°C, and that we assume that the steam is at 100°C.

Constants that we have a tendency to  need:

Heat capability of water is $\[Cpw\text{ }=\text{ }4.186J/gK\]$

Heat capability of copper is $\[Cpc\text{ }=\text{ }0.390J/gK\]$

Heat of fusion of ice $\[Hi\text{ }=\text{ }333.55J/g$

Heat of vaporization of water $\[Hw\text{ }=\text{ }2257J/g\]$

1. Melt the ice: $\[\Delta H1\text{ }=\text{ }m\left( Hi \right)\text{ }=\text{ }100\left( 333.55 \right)\text{ }=\text{ }33,355J$

We currently have 600g of water at 0°C,

2. Raise the water to 40°C:

$\Delta H2\text{ }=\text{ }mCpw\Delta T\text{ }=\text{ }600\left( 4.186 \right)\left( 40 \right)\text{ }=\text{ }1,00,464J$

3. Raise the copper to 40°C: $\[\Delta H3\text{ }=\text{ }mCpc\Delta T\text{ }=\text{ }200\left( 0.390 \right)\left( 40 \right)\text{ }=\text{ }3120J$

4. Add steam $\Delta H4\text{ }=\text{ }x\left( Hw \right)\text{ }=\text{ }2257x$

Now we know that $\[\Delta H4\text{ }=\text{ }\Delta H1\text{ }+\text{ }\Delta H2\text{ }+\text{ }\Delta H3\]$

So,

$\[2257x\text{ }=\text{ }33,355\text{ }+1,00,464J+3120J$

Solving the above expression, we get

$& 2257x\text{ }=\text{ }1,36,939 \\ \\ & x=1,36,939/2257 \\ & \\ x=\text{ }608.62 \end{array}\]$

Therefore, we need to add $\[608.62g\]$ of steam at 100°C.