MC08 ourses Specific Heat C)4 of5 e Home Constants 1 Petlodic Table Part A The h
ID: 1040194 • Letter: M
Question
MC08 ourses Specific Heat C)4 of5 e Home Constants 1 Petlodic Table Part A 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 A volume of 90.0 mL of H20 is initially at room temperature (2200 C) A chilled steel rod at 2.00 Cis placed in the water If the final temperature of the system is 21.50 °C,what is the mass of the steel bar 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 Use the following values Area specific heat of water 4.18 J/(eC) specific heat of steel 0 452 J/C) Note the diference between the terms molar heat e Materials capacity, which has units of J/(mol.°C), and Express your answer to three significant figures and include the appropriate units. View Available Hintis) specific heat, which has units of J/(g.c) mass of the steel- Submit Part B MacBook ProExplanation / Answer
mass of water = 90 g
specific heat = 4.184 J / goc
temperature = 22.00 oC
temperature of steel = 2.00 oC
final temperature = 21.50 oC
heat loss by water = heat gain by steel rod
(m Cp dT)water = (m Cp dT)steel
90 x 4.184 x (22 - 21.50) = m x 0.452 x (21.50 - 2)
m = 21.34 g
mass of steel = 21.3 g
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