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3 years ago

How much would the temperature of 275 g of water increase if 36.5 kj of heat were added?

Chemistry

2 answers:

Andrew [12]3 years ago

8 0

<u>Answer:</u> The temperature increase will be 31.70°C.

<u>Explanation:</u>

To calculate the increase in the temperature of the system, we use the equation:

$q=mc\Delta T$

where,

q = Heat absorbed = 36.5 kJ = 36500J

m = Mass of water = 275 g

c = Specific heat capacity of water = $4.186J/g^oC$

$\Delta T$ = change in temperature = ? °C

Putting values in above equation, we get:

$36500J=275\times 4.186J/g^oC\times \Delta T\\\\\Delta T=31.70^oC$

Hence, the temperature increase will be 31.70°C.

BlackZzzverrR [31]3 years ago

8 0

<span>The solution is found using the formula change in Temperature = heat supplied/ (mass of substance x specific heat). The solution would be change in temperature = 36.5kJ/(275 g x 4.184 J/g) = 31.7 degrees.</span>

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Read 2 more answers

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<h3>Answer:</h3>

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<h3>General Formulas and Concepts:</h3>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Chemistry</u>

<u>Atomic Structure</u>

Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

0.55 mol Cu(NO₃)₂

<u>Step 2: Identify Conversions</u>

Avogadro's Number

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 0.55 \ mol \ Cu(NO_3)_2(\frac{6.022 \cdot 10^{23} \ molecules \ Cu(NO_3)_2}{1 \ mol \ Cu(NO_3)_2})$
[DA] Multiply/Divide [Cancel out units]: $\displaystyle 3.3121 \cdot 10^{23} \ molecules \ Cu(NO_3)_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

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