What is the change in entropy when `2.5` mole of water is heated from `27^(@)C` to `87^(@)C`? Assume that the heat capacity is constant `(C_(p))_(m)(H_(2)O)=4.2J//g=k,ln(1.2)=0.18)`
A sample of 100 g water is slowly heated from 27^((0))C to 87^((0))C Calculate the water =4200 J kg ^(-1)K^(-1) .
Find the increase in mass when 1 kg of water is heated from 0^@C to 100^@C . Specific heat capacity of water = 4200 J kg^(-1) K^(-1) .
Calculate the change in internal energy when 5g of air is heated from internal energy when 5g of air is heated from 0^(@)C "to" 2^(@)C . Specific heat of air at constant volume is 0.172 cal//g//^(@)C .
A sample of 100 g water is slowly heated from 27^@C to 87^@C . Calculate the change in the entropy of the water specific heat capacity of water =4200(J)/(kg-K)
Calculate the entropy change when 1 kg of water is heated from 27^(@)C to 200^(@)C forming supper heated steam under constant pressure. Given specific heat of water = 4180 J Kg^(-1)K^(-1) and specific heat of steam = 1670 + 0.49 J kg^(-1)K^(-1) and latent heat of vaporisation =23 xx 10^(5) J kg^(-1) .
At constant pressure 200g of water is heated from 10^(@)C to 20^(@)C .Thus,increase in entropy is (molar heat capacity of water at constant pressure is 75.3JK^(-1)mol^(-1) )
The temperature of 2 moles of a gas is changed from 20^(@)C to 30^(@)C when heated at constant volume. If the molar heat capacity at constant volume is 8 J mol^(-1)K^(-1) , the change in internal energy is
NARENDRA AWASTHI-THERMODYNAMICS-Level 3 - Match The Column
