`420 J` শক্তি `10g` জলে সরবরাহ করলে এর তাপমাত্রা বৃদ্ধি পাবে

420 J of energy supplied to 10g of water will rises its temperature by

G4T, J 10 R, M 20 P, P 43 N, S 90L

The degree of dissociation is 0.4 at 420 K and 1.0 atm for the gaseous reaction PCl_(5)(g)rarrPCl_(3)(g)+Cl_(2)(g) . Assuming ideal behaviour of all gases, calculate the density of equilibrium mixture at 420 K and 1.0 atm. (P = 31, Cl = 35.3)

In a mixture of 35 g of ice and 35 g of water in equilibrium, 4 gm steam is passed. The whole mixture is in a copper calorimeter of mass 50 g . Find the equilibrium temperature of the mixture. Given that specific heat of water is 4200 J/kg K and that of copper is 420 J/Kg K and latent heat of fusion of ice is 3.36xx10^(5) J/kg and latent heat of vaporization of water is 2.25xx10^(6) J/kg.

If 10 g of ice is added to 40 g of water at 15^(@)C , then the temperature of the mixture is (specific heat of water = 4.2 xx 10^(3) j kg^(-1) K^(-1) , Latent heat of fusion of ice = 3.36 xx 10^(5) j kg^(-1) )

The specific heat of a certain substance is 0.86 J g^(-1) K^(-1) . Assuming ideal solution behavior. The energy required (in J) to heat 10 g of 1 molal of its aqueous solution from 300 K is closed to : [Given molar mass of the substance = 58 g "mol"^(-1) , specific heat of water = 4.2 J g^(-1) K^(-1) ]