A system absorbs 300 cal of heat. The work done by the system is 200 cal. `Delta U` for the above change is

A gaseous system absorbs 450 cal heat. The work done by the system is 836.2 J. Find the change in internal energy of the gas in calorie. (J=4.186 J "cal"^(-1) )

A system absorbs 300cal of heat , its volume doubles and temperature rises from 273 to 298 k , the work done on the surrounding is 200 cal . Delta E for the above reaction is :

A system absorbs 300cal of heat , its volume doubles and temperature rises from 273 to 298 k , the work done on the surrounding is 200 cal . Delta E for the above reaction is :

a thermodynamic system is taken from the state i to statef along the path iaf. It is found that the heat absorbed by the system is 40 cal and work done by the system is 16.19 cal. Along the path i bf, the heat absorbed is 35 cal. (a) Find the work done along the path i bf. (b) If W = - 10 cal for the curved return path fi, what is for this path ? (c) Take U_(b) = 20 cal, what is U _(f)? (d) If U _(b) = 20 cal, what is for the process ib?

In a process, q cal heat is absorbed by a closed system. If work done by the system is w cal, then what will be the value of the change in internal energy (DeltaU) in this process?

A system absorbs 920 J of heat and does work. The change in internal energy (Delta U) for the process is 460 J. The work done by the system is

A system is provided with 200 cal of heat and the work done by the system on the surrounding is 40J . Then its internal energy

A system is provided with 200 cal of heat and the work done by the system on the surrounding is 40 j. Then its internal energy

A system absorbs 115J heat from the surroundings and 290 J work is done on the system calculate the change in internal energy of the system.

A system absorbs 920 J of heat and does work. The change in internal energy (DeltaU) for the process is 460 J. The work done by the system is ________ .