Question

Review Constants Periodic Table The heat capacity of an object indicates how much energy that object can absorb for a given increase in that object's temperature. In a system in which two objects of different temperatures come into contact with one another, the warmer object will cool and the cooler object will warm up until the system is at a single equilibrium temperature. In this problem answers are requested to three significant digits for grading purposes. The true number of significant digits may be more or less. Part A Note the difference between the terms molar heat capacity, which has units of J/(mol. °C), and specific heat, which has units of J/(g."C). A volume of 80.0 mL of H2O is initially at room temperature (22.00C). A chilled steel rod at 2.00 °C is placed in the water. If the final temperature of the system is 21.20 °C, what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g. C) specific heat of steel = 0.452 J/(g. C) Express your answer to three significant figures and include the appropriate units. View Available Hint(s) ? HA Value O 2 Units mass of the steel - Submit
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Answer 1

Part A A chilled steel rod (2.00°C) (Cota) equivalent to Done Hot 22.00°C (hot) symbol for hot in h and cold & If = final temperature = 21.20°c mase of steel Bular H₂O → CPM - 4:187 |(9, 0c) Steel Cpe = 0.452 J/g.°c) heat given By H₂O = heat taken by steel Bar antilon (TS-Tn) = mclpo (Tf - Tc) -nu 4-187 (01:20-22:00): = n c 0.4527 (21.20 - 2.00)°C gic gooo – 80 5f *4,187 (-0.804) = ne 0:4827 ( 19.2%) AY g ne = 80 X4:18 X 0.8 0,452 x 19.2 Mc = 30.8259 g mass of steel Bar = 30.82599

Past B specific heat capacity of water = 4.185 / 9.00 Molas heat capacity = ? Molar mass of H₂O = 2x1 + 16 - 184/mol Molces heat capacity of water = 4.185 X '184 8.00 mol = (4.18 X18) T mol °C - 75.240 T / moloc a 75.24 Il moloc