Question

Jor General, Orgmic, and Biological Chemistry Calculations 8. AT(C) for water (7-6) First run Second run 9. Mass of water (4-3) 52.11 10. Heat (J) gained by water (Show calculations.) TeMp Chonye1-4.87 Temp 5213.91 X41 CNORC 2-3-54C Heat Gan tor waker 9XATXSHH2) 8.4134 -184 11. Heat (J) lost by the metal Heat (J) lost by metal Heat (J) gained by water (10) Th-01O 76-590 13. Specific heat of the metal g "C)3573 27654 9863 12. AT (C) for metal (5-7) (Show calculations.) 5ו רר 27 SACmelal) 276 14. Identity of metal Z2nt d cudala Questions and Problems Cad 16 a2inc coai Q3 Why did you need to transfer the metal quickly from the hot water bath to the water in the cup calorimetert neadel In be travreved to thc water in Hhe cup so tort yo wuich4 tose the heat of tHhe meta tl the cabler aw in tie envivoment-T oud maKe the Youts in aclurak Q4 How many calories are required to raise the temperature of 225 g of water from 42 °C to 75 °C? os A metal object with a mass of 19 g is heated to 96 C and then transferred to a calorimeter containing 75 g of water at 18 °C. If the water and metal object reach a final temperature of 22 °C, what is the specific heat of this metal object? 2761

Answer 1

1 04) Ciren mass of water = 2059, temperature laned funder - 175 -423 AT: 350 Heat Capacity of water is I Calzie per gran dege. now & Amout of head (@) = mSAT = 2257 1 Calbic x 33 c = 7425 20 Hele S = Specific ps). Quantity of head (6)= mSAT Some heat is Lost by the metal object & that the equal to heat gained by water So now heat lost by metal object = msot I head of the metal = 19xSx (96-22) - 1466x S J-0 now heat gowned by watel = ms st (Here se specifur heal gwater) = 95 x 4.186x (22-18) 4.186 I = 1255.850 Si specific heat of the metal = 0.893 J/goc. or equating 0 8