Question

2 kg of air is heatef at constant volume.the temperature of air is increased from 293k to 313k if the specific heat of air at constant volume is 0.718kj/kh k,the amount of heat absorbed in kj and kcal is

Answer 1

Heat absorbed H = m s T
H = 2 kg * 0.718 kJ/ °K/kg * (313-293)°K
H = 28.72 kJ
H = 28.72/4.18 kCal

Answer 2

Q = mSdT
= 2 kg × 0.718 kJ/(kg K) × (313 K - 293 K)
= 28.72 kJ
Amount of heat absorbed is 28.72 kJ

To covert kJ to kcal you should know that
1 kcal = 4.184 kJ

So,
28.72 kJ = 28.72 kJ × 1 kcal/(4.184 kJ)
= 6.864 kcal