A hot 102.4 g lump of an unknown substance initially at168.9 °C is placed in 35.0 mL of water initially at 25.0 °C andSubstanceSpecific heat (J/(g-°C))aluminum0.897the system is allowed to reach thermal equilibrium. The finaltemperature of the system is 55.7 °C.graphite0.709rhodium0.243titanium0.523tungsten0.132Using this information and the specific heat values forzinc0.388several metals in the table, identify the unknownwater4.184substance. Assume no heat is lost to the surroundings.O tungstenO graphiterhodiumtitaniumO aluminumO zincétvWMacBook AirDII80888F10F11F12FBF9F6F7F3F4F524%3Ddelete

Given: Temperature of unknown metal decreases from 168.9 °C to 55.7 °C Temperature of water…

A hot 102.4 g lump of an unknown substance initially at 168.9 °C is placed in 35.0 mL of water initially at 25.0 °C and Substance Specific heat (J/(g:°C)) aluminum 0.897 the system is allowed to reach thermal equilibrium. The final temperature of the system is 55.7 °C. graphite 0.709 rhodium 0.243 titanium 0.523 Using this information and the specific heat values for tungsten 0.132 zinc 0.388 several metals in the table, identify the unknown water 4.184 substance. Assume no heat is lost to the surroundings. O tungsten graphite O rhodium titanium O aluminum O zinc

A hot 102.4 g lump of an unknown substance initially at 168.9 °C is placed in 35.0 mL of water initially at 25.0 °C and Substance Specific heat (J/(g:°C)) aluminum 0.897 the system is allowed to reach thermal equilibrium. The final temperature of the system is 55.7 °C. graphite 0.709 rhodium 0.243 titanium 0.523 Using this information and the specific heat values for tungsten 0.132 zinc 0.388 several metals in the table, identify the unknown water 4.184 substance. Assume no heat is lost to the surroundings. O tungsten graphite O rhodium titanium O aluminum O zinc

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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SUBJECT : GENERAL CHEMISTRY 2 TOPIC : HEAT AND CALORIMETRY
A 46.9-g iron rod, initially at 18.6 oC is submerged into 50.4 g of water at an unknown initial temperature. The final temperature of the mixture upon reaching thermal equilibrium is 69.6 oC. What is the initial temperature of the water (in oC)? The specific heat capacity of iron is 0.449 J/g.oC The specific heat capacity of water is 4.184 J/g.oC
The table lists the specific heat capacities of select substances. Substance Specific heat capacity (J/g °C) lead 0.128 silver 0.235 copper 0.385 iron 0.449 aluminum 0.903 water 4.184 An unknown substance has a mass of 19.5 g.19.5 g. The temperature of the substance increases by 16.1 ∘C16.1⁢ ∘C when 73.8 J73.8 J of heat is added to the substance. What is the most likely identity of the substance?
A calorimeter contained 67.0 g of water at 15.93°C. A 110.-g sample of iron at 66.42°C was placed in it, giving a final temperature of 19.42°C for the system. Calculate the heat capacity of the calorimeter. Specific heats are 4.184 J/g-°C for H2O and 0.444 J/g-°C for Fe. Heat capacity of the calorimeter =
The temperature of a sample of iron increased by 22.0 °C22.0 °C when 269 J269 J of heat was applied. What is the mass of the sample? Substance Specific heat J/(g · °C) lead 0.128 silver 0.235 copper 0.385 iron 0.449 aluminum 0.903
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A 4.70-g sample of aluminum pellets (specific heat capacity = 0.89 J °C- g) and a 10.00-g sample of iron pellets (specific heat capacity = 0.45 J °C- g) are heated to 115.0°C. The mixture of hot iron and aluminum is then dropped into 98.3 g of water at 22.0°C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings. The specific heat capacity of water is 4.18 J °C¯'g¬! °C
The molar heat of solution of a substance is found to be +21.38 kJ/mol. The addition of 0.100 mol of this substance to 1.000L of water initially at 40.0 degrees celsius results in a temperature decrease. Assume the specific heat of the resulting solution to be equal to that of pure water. Find the final temperature of the solution (Also assume that the heat capacity of the calorimeter is negligible).
When a 45.0 g sample of an alloy at 100.0 °C is dropped into 100.0 g of water at 25.0 °C, the final temperature is 37.0 °C. What is the specific heat of the alloy? The specific heat of H20(1) is 4.184 J'g-1.°c-1. q = mc AT O A. 0.423 J-g-1.°c-1 O B. 9.88 J-g-1.°C-1 O C. 48.8 J'g-1.°C-1 O D. 1.77 Jg-1.°c-1
A 14.7-g sample of iron pellets (specific heat capacity 0.460 J/°C.g) and a 13.4-g sample of gold pellets (specific heat capacity 0.126 J/ °C g) were heated to 195°C. The mixture of hot metal is dropped into 125 g of ethylene glycol (specific heat capacity 2.43 J/°C g) at 21.3°C. Assuming no heat loss to the surroundings, calculate the final temperature of the metal and ethylene glycol mixture. 1 4 7 +/- °C 2 3 5 6 8 9 0 X C x 100 C G
A beaker contained 65.35 g of water at 13.85°C. A 105.41-g sample of metal at 54.71°C was placed in it, giving a final temperature of 32.58°C for the system. Calculate the specific heat of the metal. Specific heat is 4.184 J/g/°C for H2O.
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