Question

I final = 100 u You cool a 100.0 g slug of red-hot iron (temperature 745°C) by dropping it into an insulated cup of negligible mass containing 75.0 g of water at 20.0°C. Assume no heat exchange with the surroundings. Submit Previous Answers ✓ Correct Part B What is the final mass of the iron and the remaining water? Express your answer with the appropriate units. HÅR 2 ? Value Units | m= m = Submit Request Answer

Answer 1

LetQ; be the amount of heat lost by by red hotiron piece while reaching final equlibrium temperature T, Q: =m S; (T. -T), m, Eiron mass=100e - 3 kg, iron specific heat=S; = 440J/kg. K T. = initial temperature=745°C T =100°C finlal temperature (we know from part A) weplug the valuesin hand, while using the relation 0x =28600J LetQ, be the amount of heat gained by water while reaching final equlibrium temperature T=100°C Q. =M_S(T. -T)+m L M = total mass of liquid water= 75e-03 kg Sv=water specific heat=4200J/kg C m. =mass of water evoparated T =20°C L. = 2.23e+00 home we plug the valuesin hand, while using the relation by the conservation of energy Qu=Q;=M_S(T - T)+m L,; we plug the know valuesin the relation gives =>m = 0.001525 kg we plug the know values in the relation gives final mass of water remaining =M,-mv = 73.48g