How many mL of water are present if 1300 KJ is given off as energy when water is

Question

How many mL of water are present if 1300 KJ is given off as energy when water is cooled from 373.5 K to 295.2 K (Given: the specific heat of water is 4.184 J/gK and density of water is 1 g/mL) g = m C_s delta T g = T_F - T; If 44.28 g of Al at 23.0 degree C is placed in 90.0 g of H_2 O at 79.0 degree C, what is the final temperature of the mixture in degree C? (Specific heat capacity of Al = 0.900 J/g degree C) You place 15.67 g of a metal into boiling water and let it reach equilibrium at 98.7 degrees Celsius. You then place the metal into 75 mL of water, which is set at a temperature of 24.5 degrees Celsius. When thermal equilibrium is reached, the final temperature is 27.9 degrees Celsius. What is the specific heat of the metal? (Given: specific heat of water is 4.184 J/gk and density of water is 1g/mL) Write the balanced chemical equation for the formation of COCl_2 (g) from the natural elements and look up the value for delta Hr^0 in the Appendix in the back of the book to determine the enthalpy of the reaction.

Explanation / Answer

Questio 17.

Using the expression Q = mCp(Tf-Ti) you can find the mass, and then the volume:

m = Q/Cp(Tf-Ti)
m = 1300000 / 4.184(373.5-295.2)
m = 3968.17 g

The density of water is 1 g/mL. So this mass is the same as volume: V = 3968.17 mL or 3.968 L

Question 18.

For Al: Q1 = 44.28*0.9(Tf-23) = 39.852Tf - 916.596
For water: Q2 = 90*4.184(79-Tf) = 29748.24 - 376.56Tf

Q1 = Q2 then:
39.852Tf - 916.596 = 29748.24 - 376.56Tf
39.852Tf + 376.56Tf = 29748.24 + 916.596
416.412Tf = 30664.84
Tf = 73.64 ºC

The other two questions post them in another question thread. However question 19 is solved in a similar way as this question, only that instead of solving for Tf you solve for cp.

Hope this helps

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