Question
Ty
please explain
When a hollow object is heated, what happen to its inside and outside dimension? The outside and inside dimensions both get smaller. The outside and inside dimensions both get larger. The outside dimension gets larger and the inside dimension gets smaller The outside dimension gets smaller and the inside dimension gels larger. The relative change in inner compound to outer dimensions depends upon the precise relation between the coefficients of linear and volume expansion. a heating coil inside an electric kettle consumes 1 5 kW of electrical power How long will if take to raise the temperature of 0. 50 kg of water (about one pint) from 20 degree C to 98 degree C? Neglect the change in thermal energy of the coil and kettle and assume no heat is lost to the environment. 20 s 110 s 160 s 420s 800 660 s a diatomic ideal gas cools at a constant pressure of 6. 0 x 104 Pa from a temperature of 27 degree C to 33 degree C. The heat energy given off by the gas in this process is 4200 J. The initial volume was 0. 10 m3. The internal energy of the gas decreases in this process by: 5400 J 4800 J 4200 J 3600 J 3000 J In the mechanical equivalent of heat experiment, the graph of number of turns versus temperature is a perfectly straight line if the turning rate is kept constant. a curve whose slope increases slightly with increasing temperature because the cylinder can absorb less heat for a given temperature rise as its temperature increases. a perfectly straight line because the energy extracted from the suspended mass is independent of the turning rate. a curve whose slope increases slightly with increasing temperature because heat flows from the air to the cylinder when the cylinder is cooler than the air in the room warmer than room air, heat flows the other way a curve whose slope increases slightly with increasing temperature because the aluminum conducts heat less effectively as the temperature rises.
Explanation / Answer
4 ) d , because inside we get compressed and ouside will get elonget
5) time = m*C*(T2-T1)/(1.5k) =0.5(98-20)*4200/(1.5k) =109.2 s = (B)
6) n = number of moles = PV/(RT) = 6*10^4*0.1/(8.31*300)= 2.4
change in internal energy = n*(5/2)R(-33-27) = - 2988 j = (E)
(7) a
8) heat = m*s=0.2*3.36*10^5 = 0.67*10^5 J
heat removed rate = 0.67*10^5/(30*60) =37.33 w = (G)
