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10 g of ice at 0°C is mixed with 100 g of water at 50°C. What is the resultant temperature of the mixture? (Given: Specific heat of water = 1 cal/gram/°C; Latent heat of ice = 80 cal/gram)
Let the final temperature of the mixture be T. Heat lost by water = Heat gained by ice
During the adiabatic expansion of 2 moles of a gas, the internal energy was found to have decreased by 100 J. The work done by the gas in this process is
dU = -100 J dQ = dU + dW dQ = 0 (Adiabatic process) dW = -dU = +100 J
For a diatomic gas, change in internal energy for a unit change in temperature for constant pressure and constant volume is U1 and U2, respectively. What is the ratio of U1 to U2?
In case of the diatomic gas the internal energy is given by U = 5/2 nRT As the temperature of the both gases is same. Hence U1 : U2 :: 1 : 1
The work of 146 kJ is performed in order to compress one kilo mole of a gas adiabatically, and in this process the temperature of the gas increases by 70C. The gas is (R = 8.3 J mol-1 K-1)
Work done in an adiabatic process
The temperature-entropy diagram of a reversible engine cycle is given in the figure. Its efficiency is
A polyatomic gas (γ = 4/3) at pressure P is compressed to (1/8)th of its initial volume adiabatically. The pressure will change to
Calculate the temp. in K at which a perfect black body radiates at the rate of 5.67 watt cm-2, σ = 5.67 ×10-8 watt m-2 K - 4
Rate of energy dissipated per unit are per second is given by
T4 = 1012 ⇒ T = 1000 K
Find the actual mechanical advantage of a machine, which has a velocity ratio of 3.2 and efficiency 75%.
Ideal mechanical advantage (IMA) = Velocity ratio = 3.2 and Efficiency = Actual mechanical advantage (AMA) / Ideal mechanical advantage(IMA) => 75/100 = AMA / 3.2 ∴ AMA = 2.4
A heat engine takes in 900 J of heat from a high temperature reservoir and produces 300 J of work in each cycle. What is its efficiency?
If a system changes from state (P1, V1) to (P2, V2) as shown below, the work done by the system is
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