Question

Question: A 434 g steel bucket contains 8.55 L of water at 13.3 °C. The bucket and water are in thermal equilibrium with each other. Part 1) How much energy needs to be added to the bucket to raise its (and its contents) temperature by 56.0 °C? Q= Part 2) A nugget of lead at 437 °C with a mass of 697 g is now added to the bucket (while it is at 13.3°C). Once the system has reached thermal equilibrium, what will the temperature of the water be? Assume no energy is lost or gained from the surrounding. TE C Part 3) Now assume that as the nugget is added 3.34 g of water is converted to steam. If no other energy is lost, what is the final temperature of the system? T Check

Specfic heat of Steel is 456

Answer 1

water- the bucket Temperature of =434 g Mass of steed bucket am, ature of 286.9 y ento 8.55 kg (m2) Temperature = 13.3°C = 286.3 (+2) Mass of water- specific heat of steel 456 -5 lugy (61) specific heat w water=4179.6-5/egy of Part 1) Energys needed = H needed = H = m₂, s. (T2 -T.) = 48406 456x (329–28613) 8450.50 J g 8450.50J 692 1000 olegy ng Mans (Ans) [5ce = 3296, 434g= 434/1000 ug] Part 2) Specific heat of lead =53= 128 of lead=m3=697g = Temperature of lead = 4.-437°c 710 at Let the final temperatur rature of the system thermal equi librium Then, m3.53.(t-1) = mins, . (+-+1) + m2.52.(T-tz) mz.s3.tamisit, tm 2b2t2 mibot m2b2t masa (0.697x128*7109 +(0.43484864 286-3)+(8.55* 417906 X 286.3 (0.4344456) 4(8.55x4139.6)+(0.697x128) be T. T =

= 287.35k 14.35 °C T = 14.35 °C (Ang) Part 3) (L) --- Latent heat of Steam= 2257hs /ng final temperature of the system will be Leto the Th. Then a m3.53.(t-1') = m, 5,. (I'-to) + m2-3.34.52:47 100 + 3.34 XL 1000 - 3.34*L 1000 = > misit, & (M2 -3.34) . b2ctz T'. 1000 mis, at m2-3.34,62 + Mzs3 1000 - 285.38K 12.38°c T= 12.38°e (ams) -