Calculate the gas constant for 1 g of gas from the following data :
`C_p=0.245 "cal g"^(-1).^@C^(-1), C_v=0.165" cal g"^(-1).^@C^(-1)and J=4.2 xx10^7 erg cal^(-1)`

Calculate the velocity with which a piece of ice at - 5^(@)C be thrown against a wall so that its entire mass melts upon contact. Latent heat of ice = 80 cal g^(-1) , specific heat of ice = 0.5 cal g^(-1).^(@)C^(-1) . And J=4.2 xx 10^(7) erg cal^(-1) .

One cubic metre of air at 27^(@)C and 10^(5)Nm^(-2) pressure weighs 1.18kg . Calculate the value of the gas constant for 1kg of the gas and calculate c_(p) of air if 168 cal kg^(-1)K^(-1) and J=4.2Jcal^(-1)

Calculate the two specific heats of nitrogen from the following data : gamma = c_(p)//c_(v) = 1.51 density of nitrogen at N.T.P = 1.234 g litre ^(-1) and J = 4.2 xx 10^(7) erg cal^(-1) .

Find the amount of heat enegy required to convert 100 g of ice at -10^(@)C into stea at 120^(@)C . (Take S_(ice)=0.5" cal "g^(-1).^(@)C^(-1),S_(W)=1" cal "g^(-1)^(@)C^(-1),S_("Steam")=0.5" cal "g^(-1).^(@)C,L_(f)=80" cal "g^(-1),L_(V)=540" cal "g^(-1) )

An insulated container has 60 g of ice at -10^(@)C . 10 g steam at 100^(@)C , sourced from a boiler, is mixed to the ice inside the container. When thermal equilibrium was attained, the entire content of the container was liquid water at 0^(@)C . Calculate the percentage of steam (in terms of mass) that was condensed before it was fed to the container of ice. Specific heat and latent heat values are S_("ice") = 0.5 cal g^(-1) .^(@)C^(-1) , S_("water") = 1.0 cal g^(-1) .^(@)C^(-1) L_("fusion") = 80 cal g^(-1) , L_("vaporization") = 540 cal g^(-1)

Calculate the value of c_(v) for air, given that c_(p) = 0.23 calorie g^(-1)K^(-1) . Density of the air at S.T.P. is 1.293 g litre ^(-1) and J = 4.2 xx 10^(7) erg calorie ^(-1) .

A calorimeter of water equivalent 10 g contains a liquid of mass 50 g at 40 .^(@)C . When m gram of ice at -10^(@)C is put into the calorimeter and the mixture is allowed to attain equilibrium, the final temperature was found to be 20^(@)C . It is known that specific heat capacity of the liquid changes with temperature as S = (1+(theta)/(500)) cal g^(-1) .^(@)C^(-1) where theta is temperature in .^(@)C . The specific heat capacity of ice, water and the calorimeter remains constant and values are S_("ice") = 0.5 cal g^(-1) .^(@)C^(-1), S_("water") = 1.0 cal g^(-1) .^(@)C^(-1) and latent heat of fusion of ice is L_(f) = 80 cal g^(-1) . Assume no heat loss to the surrounding and calculate the value of m.

For air, specific heat at constant pressure is 0.273 cal g^(-1) .^(@)C^(-1) and specific heat at constant volume is 0.169 cal g^(-1).^(@)C^(-1) , density of air = 0.001293 g cm^(-3) at S.T.P. Calculate the value of J.

The water equivalent of 50g of aluminium (Specific heat is 0.2 cal g^-1 °C^-1 )