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AleksAgata [21]
3 years ago
7

Example 1: urea, (nh2)2co, is used in the manufacture of resins and glues. when 5.00 g of urea is dissolved in 250.0 ml of water

(d = 1.00 g/ml) at 30.0 c, 27.6 kj of heat is absorbed. (specific heat of water 4.18 j/g.c) is the solution process exothermic? what is ? what is the final temperature of solution?
Chemistry
2 answers:
katen-ka-za [31]3 years ago
7 0
A) i believe the reaction is exothermic, because 27.6 kg of thermal energy is gained by the water solution, the dissolution of urea is exothermic. Exothermic reaction is a chemical reaction that releases energy by light or heat. It is the opposite of an endothermic reaction where heat is gained by the reaction.

b) The water gained the heat released when urea dissolve. That is the water gained 27.6 kJ, while dissolution of urea released 27.6 kJ. Therefore, the heat gained is equal to the heat lost.

c) From part B, since heat gained is equal to heal lost, then
250 g × (Tf -30) ×4.18 J/g = 27600 J
 = 1045 Tf - 31350J = 27600 J
Tf = 56.41°C.
Therefore the final temperature  of the solution is 56.41°C

d) The initial and final temperature in Fahrenheit
°F = °C × (9/5) +32,
Thus, 30°C will be equal to 86° F
while 56.41 will be equivalent to  133.54 ° F


Sholpan [36]3 years ago
4 0

trampoline and playground

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Practice Problem: True Stress and Strain A cylindrical specimen of a metal alloy 49.7 mm long and 9.72 mm in diameter is stresse
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Answer:

The true stress required = 379 MPa

Explanation:

True Stress is the ratio of the internal resistive force to the instantaneous cross-sectional area of the specimen. True Strain is the natural log to the extended length after which load applied to the original length. The cold working stress – strain curve relation is as follows,

σ(t) = K (ε(t))ⁿ, σ(t) is the true stress, ε(t) is the true strain, K is the strength coefficient and n is the strain hardening exponent

True strain is given  by

Epsilon t =㏑ (l/l₀)

Substitute㏑(l/l₀) for ε(t)

σ(t) = K(㏑(l/l₀))ⁿ

Given values l₀ = 49.7mm, l =51.7mm , n =0.2 , σ(t) =379Mpa

379 x 10⁶ = K (㏑(51.7/49.7))^0.2

K = 379 x 10⁶/(㏑(51.7/49.7))^0.2

K = 723.48 MPa

Knowing the constant value would be same as the same material is being used in the second test, we can find out the true stress using the above formula replacing the value of the constant.

σ(t) = K(㏑(l/l₀))ⁿ

l₀ = 49.7mm, l = 51.7mm, n = 0.2, K = 723.48Mpa

σ(t) = 723.48 x 106 x (㏑(51.7/49.7))^0.2

σ(t) = 379 MPa

The true stress necessary to plastically elongate the specimen is 379 MPa.

6 0
3 years ago
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