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suter [353]
4 years ago
13

The statements in the tables below are about two different chemical equilibria. The symbols have their usual meaning, for exampl

e AG Gibbs free energy of reaction and stands for the equilibrium constant. stands for the standard In each table, there may be one statement that is false because it contradicts the other three statements. If you find a false statement, check the box next to It. Otherwise, check the "no false statements box under the table. statement false? statement false? Ink>0
AH° R<1
AG'>0
AG'>0
In > AH">TAS
no false statements

Chemistry
1 answer:
Talja [164]4 years ago
4 0

Answer:

see explaination

Explanation:

Please kindly check attachment for the step by step solution of the given problem

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Energy in the amount of 420 J is added to a 35 g sample of water at a temperature of 10°C. What is the final temperature of the
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The <em>final temperature </em><em>of the</em><em> water, T2</em> = <em>38.57°C</em>

Temperature can be defined as a measure of the degree of hotness or coldness of a physical object (body). Thus, it is measured with a thermometer and its units are degree Celsius (°C), Fahrenheit (°F) and Kelvin (°K).

A calorie refers to the amount of heat required to raise the temperature of a gram of water by one (1) degree Celsius (1°C).

<u />

<u>Given the following data:</u>

  • Quantity of energy = 420J
  • Mass = 35 grams
  • Initial temperature, T1 = 10°C

The specific heat capacity of water is 4.2 J/g°C.

To find the final temperature of the water (T2):

Mathematically, the quantity of energy (heat capacity) is given by the formula;

Q = mcdt

Where;

  • Q represents the heat capacity or quantity of heat.
  • M represents the mass of an object.
  • C represents the specific heat capacity of water.
  • dt represents the change in temperature.

Substituting the values into the formula, we have;

420 = 3.5 \; * \; 4.2 \; * \;  dt

420 = 14.7 \; * \; dt\\\\dt = \frac{420}{14.7}

Change in temperature, dt = 28.57°C

Next, we would solve for the final temperature by using this formula;

dt = T2 - T1

28.57 = T_{2} - 10\\\\T_{2} = 28.57 \; + \; 10\\\\T_{2} = 38.57

<em>Final temperature, T2 = 38.57°C</em>

<em>Therefore</em><em>, </em><em>the</em><em> final temperature </em><em>of the</em><em> water, T2</em> is equal to <em>38.57°C</em>

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