Use Hess\'s Law to find delta H for the reaction: A 466-gram sample of water it

Question

Use Hess's Law to find delta H for the reaction: A 466-gram sample of water it bearded from 8.50 degree C to 74.60 degree C. Calculate the amount of heal absorbed by the water. The specific heat of water is 4.184 J/g. degree C. A 53.0-gnm aluminum block initially at 27.5 degree C absorbs 725 j of heal. What is the final temperature of the aluminum? The specific heat of aluminum it 0.903 J/g-degree C. Suppose you find a pre-1982 penny in the mow. How much heal is absorbed by the penny as it warms from the temperature of the mow (-8 degree C) to the temperature of your body. 37degree C. Assume the penny is pure copper and has a mass of 3 10 grams. The specific heal of Cu is 0.385 J/g- degree C. A lead pellet having a mas of 26.47 grams at 89.98 degree C is dropped into 100.0 grams of water in an insulated. The temperature rose from 22.50 degree C to 23.17 degree C. What is the specific heal of the lead pellet? A 465-grms of iron is removed from an oven and dropped into 375.0 grams of water in an insulated container. The temperature of the increases from 26.0 to 87.0 degree C. If the specific heal of iron is 0.449 J/g-degree C, what have been original oven temperature? Assume all the heat was transferred from the iron to the water and none to the surroundings. Consider the ether CH_3CH_2O-C(CH_3)_3(t). Write the combustion reaction involving this ether. Write the formation reaction involving this ether Consider the alcohol HO-C(CH_3)_3(f) Write the combustion reaction involving this alcohol. Write the formation reaction involving this alcohol. Consider the ketone (CH_3)_2C=O(l). Write the combustion reaction involving this ketone. Write the formation reaction involving this ketone. Consider the carboxylic acid C_6H_5COOH(s). Write the combustion reaction involving this carboxylic acid. Write the formation reaction involving this carboxylic acid.

Explanation / Answer

Question 14

Q = m c T
Q = quantity of heat in joules (J)
m = mass of the substance acting as the environment in
grams (g) = 26.47
c = specific heat capacity in J/(g oC) = it is for water 4.18 J/(g oC)
T = change in temperature = Tfinal - Tinitial in oC

specific heat (c) = Heat Energy / (mass of substance * change in temperature)

The heat lost by lead = the heat gained by water

26.47 x Sp heat x (89.98 -23.17) = 100 x 4.18 x (23.17 - 22.5)

Sp heat of lead = 0.1583 J/(g oC)

Question 15

Use the same formula

heat lost by iron = heat gained by water

465 x 0.449 x (Tf - 87) = 375 x 4.18 x ( 87 - 26 )

208.78 Tf - 18164.295 = 95617.2

208.78 Tf = 113781.795

Tf = 544.9

Hence oven temperature must be 544.9 Deg Celcius

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