Question

In a laboratory experiment for finding the specific latent heat of ice; 100 g of water at 30 degree celsius was taken in a calorimeter made of copper and mass 10g . When 10g of ice at 0 degree celsius was added to the mixture and kept within the liquid till the ice melted completely , final temperature of mixture was found out to be 20 degree celsius .
1. What is the total quantity of water in a calorimeter at 20 degree celsius?
2. Specific Heat Capacity of water being 4.2J/g/C and 0.4 J/g/C respectively, what quantity would each release in cooling down to 20 degree celsius from the initial stage.
3. Write an expression for heat gained by ice on melting.
4. Calculate the value of latent heat of fusion of ice from the data discussed above.

Answer 1

Answer:

Latent heat of the ice is 382 J/g obtained from above experimental data

Explanation:

1)

As we know that whole ice will melt down at 20 degree celcius

So final amount of water at this temperature is given as

$m = 100 g + 10 g$

$m = 110 g$

2)

Heat released by water + calorimeter is given as

$Q = m_1s_1\Delta T + m_2s_2\Delta T$

so we will have

$Q = 100(4.2)(30 - 20) + 10(0.4)(30 - 20)$

$Q = 4200 + 40$

$Q = 4240 J$

3)

Heat gain by the ice = heat released by water + calorimeter

so we will have

$Q = mL + ms\Delta T'$

$4240 = 10 L + 10(s)(20 - 0)$

4)

Now from above equation we will have

$4240 = 10L + 10(2.1)(20)$

$424 = L + 42$

$L = 382 J/g$

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Topic : Calorimetery

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Answer 2

Answer:

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