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A sample of 100 g water is slowly heated...

A sample of 100 g water is slowly heated from `27^(o)C` to `87^(o)C`. Calculate the change in the entropy of the water. specific heat capacity of water =4200 j/kg k .

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To calculate the change in entropy of the water as it is heated from \(27^{\circ}C\) to \(87^{\circ}C\), we can follow these steps: ### Step 1: Convert temperatures to Kelvin First, we need to convert the temperatures from Celsius to Kelvin since the entropy formula requires absolute temperatures. \[ T_1 = 27^{\circ}C + 273.15 = 300.15 \, K \] ...
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A sample of 10g H_(2)O is slowly heated from 27^(@)C to 87^(@)C . Calculate the change in entropy during heating. (specific heat of water =4200 J kg^(-1)K^(-1)) .

500 kg of water is heated from 20^(@) to 100^(@) C . Calculate the increase in the mass of water.Given specific heat of water =4.2 xx 10^(3) J kg ^(-1) .^(@)C^(-1) .

Calculate the increase in the internal energy of 10 g of water when it is heated from 0^(0)C to 100^(0)C and converted into steam at 100 kPa. The density of steam =0.6 kg m^(-3) specific heat capacity of water =4200 J kg^(-1 ^(0)C^(-3) latent heat of vaporization of water =2.25xx10^(6) J kg^(-1)

Calculate the increase in the internal energy of 10 g of water when it is heated from 0^(0)C to 100^(0)C and converted into steam at 100 kPa . The density of steam =0.6 kg m^(-3) , specific heat capacity of water =4200 J//kgC ,latent heat of vaporization of water =2.25xx10^6 J kg^(-1)

A hot solid of mass 60 g at 100^(@)C is placed in 150 g of water at 20^(@)C . The final steady temperature recorded is 25^(@)C . Calculate the specific heat capacity of the solid. [Specific heat capacity of water = 4200 kg^(-1)""^(@)C^(-1) ]

A piece of ice of mass 40 g is added to 200 g of water at 50^@ C. Calculate the final temperature of water when all the ice has melted. Specific heat capacity of water = 4200 J kg^(-1) K^(-1) and specific latent heat of fusion of ice = 336 xx 10^3" J "kg^(-1) .

200 g of hot water at 80^@ C is added to 300 g of cold water at 10^@ C. Neglecting the heat taken by the container, calculate the final temperature of the mixture of water. Specific heat capacity of water = 4200 J kg^(-1)K^(-1)

A refrigerator converts 100 g of water at 20^@ C to ice at - 10^@ C in 73.5 min. Calculate the average rate of heat extraction in watt. The specific heat capacity of water is 4.2 J g^(-1) K^(-1) , specific latent heat of ice is 336 J g^(-1) and the specific heat capacity of ice is 2.1 J g^(-1) K^(-1)

HC VERMA ENGLISH-LAWS OF THERMODYNAMICS-All Questions
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