Question

1. A solid of mass 150g of 200 degree celsius is placed in 0.4kg ofwater at 20 degree celsius till a constant temparature is attained. Ifthe specific heat capacity of the solid is 0.5J/g K .Find theresulting temparature of the mixture.2. Calculate the amount of heat energy absorbed by 40g of ice at -10degree celsius to convert into water at 80 degree celsius. Given thatspecific heat capacity of ice is 2.1J/g degree celsius.3. Some hot water is added to three times its mass of cold water at 10degree celsius . The resulting temparature is found to be 20 degreecelsius, find initial temparature of the hot water.4. What mass of a liquid A of specific heat capacity of 0.8J/g K andat temparature of 40 degree celsius must be mixed with 100g of aliquid B of specific heat capacity 2.1J/g K and at 20 degree celsiusso that the final temparature becomes 20 degree celsius.

Answer 1

Firstly, let the final equilibrium temperature be T.

Now, the heat lost by the solid=the heat gained by the water.

$(0.5)(150)(200-T)=(4.2)(400)(T-20)$

On either sides I manipulated the values to get Joules/gK and g. We are measuring temperature difference so we need not convert to Kelvin.

On operating we get,

$T=\frac{648}{23.4}=27.7$


So the eqm temperature is 27.7 degrees Celsius.

The second part is a three step process, ice at -10 to ice 0 degrees Celsius,ice at 0 degrees celsius to water at 0 degrees Celsius and then water at 0 degrees Celsius to water at 80 degrees Celsius.

The energy absorbed by ice to get to 0 degrees is the energy needed to move up 10K or degrees Celsius,

$E_{1}=(10)(40)(2.1)=840 J$

the energy needed for phase change is,

$E_{2}=(320)(40)=12800J$

Here 320J/g is the latent heat of fusion of water.

$E_{3}=(80)(4.2)(40)=13440J$

Thus total energy needed is 
$q=E_{1}+E_{2}+E_{3}=27080J$

I'm too tired to do the rest xD

Answer 2

It took so much time to solve......

I am too tired.......

Hope it helped you.....

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