KVPY2013PhysicsCurrent Electricity
A 10.0 ~W electrical heater is used to heat a container filled with 0.5 ~kg of water. It is found that the temperature of the water and the container rise by 3 K in 15 ~minutes . The container is then emptied, dried, and filled with 2 ~kg of an oil. It is now observed that the same heater raises the temperature of the container-oil system by 2 K in 20 ~minutes . Assuming no other heat losses in any of the processes,
Options
- A2.5 10³ JK ⁻¹ ~kg ⁻¹
- B5.1 10³ JK ⁻¹ ~kg ⁻¹
- C3.0 10³ JK ⁻¹ ~kg ⁻¹
- D1.5 10³ JK ⁻¹ ~kg ⁻¹
Correct answer
A. 2.5 10³ JK ⁻¹ ~kg ⁻¹
Step-by-step solution
Pt = m _ w S _ w T + m _ c s _ c T 10 15 60=0.5 4200 3+ m _ c s _ c 39000=6300+ m _ c s _ c 3 ~m _ c s _ c =900 ~J / k . Now, for oil 10 20 60=2 S ₀ 2+900 212000-1800=4 ~S ₀ ~S ₀= 10200 4 =2.51 10³ ~J / kg - k