![SOLVED: 'Problem 1: An oil heater heats 100 kg per minute of oil from 35PC to 100PC in counter flow heat exchanger The average specific heat of the oil is 2.5 kl/kg– SOLVED: 'Problem 1: An oil heater heats 100 kg per minute of oil from 35PC to 100PC in counter flow heat exchanger The average specific heat of the oil is 2.5 kl/kg–](https://cdn.numerade.com/ask_previews/d9134f8f-cd5a-46b0-a452-65ecfd9e32c7_large.jpg)
SOLVED: 'Problem 1: An oil heater heats 100 kg per minute of oil from 35PC to 100PC in counter flow heat exchanger The average specific heat of the oil is 2.5 kl/kg–
![The specific heat of air at constant pressure is 1.005 kJ/kg K and the specific heat of air at constant volume is 0.718 kJ/kg K .Find the specific gas constant. The specific heat of air at constant pressure is 1.005 kJ/kg K and the specific heat of air at constant volume is 0.718 kJ/kg K .Find the specific gas constant.](https://dwes9vv9u0550.cloudfront.net/images/4294897/6a82cfe6-c07f-4f0e-8969-d8e46638d120.jpg)
The specific heat of air at constant pressure is 1.005 kJ/kg K and the specific heat of air at constant volume is 0.718 kJ/kg K .Find the specific gas constant.
![Two blocks of metal, each having a mass of 10 Kg and a specific heat of 0.4 Kj/Kg K, are... - YouTube Two blocks of metal, each having a mass of 10 Kg and a specific heat of 0.4 Kj/Kg K, are... - YouTube](https://i.ytimg.com/vi/Ws0jnY51NQI/sddefault.jpg)
Two blocks of metal, each having a mass of 10 Kg and a specific heat of 0.4 Kj/Kg K, are... - YouTube
![Determine the heat transfer, in kJ/kg, for the reversible process 1-3 shown in the Figure below. | Homework.Study.com Determine the heat transfer, in kJ/kg, for the reversible process 1-3 shown in the Figure below. | Homework.Study.com](https://homework.study.com/cimages/multimages/16/20.02.197962769642709008884.jpg)
Determine the heat transfer, in kJ/kg, for the reversible process 1-3 shown in the Figure below. | Homework.Study.com
![SOLVED: (a) The change in kinetic energy It is expressed as ViK ke = 2 50' 80? kJkg 2 1000 Ike =-4.95 kJkg] How come it is not Ke m flow rate SOLVED: (a) The change in kinetic energy It is expressed as ViK ke = 2 50' 80? kJkg 2 1000 Ike =-4.95 kJkg] How come it is not Ke m flow rate](https://cdn.numerade.com/ask_images/938394f908054cdcbc94e55c94cc355c.jpg)