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Maru [420]
3 years ago
6

A kettle of water is at 14.5°C. Its temperature is then raised to 50.0°C by supplying it with 5,680 joules of heat. The specific

heat capacity of water is 4.186 joules/gram degree Celsius. What is the mass of water in the kettle? Express your answer to three significant figures. The mass of the water in the kettle is grams.
Chemistry
2 answers:
Tamiku [17]3 years ago
4 0

Answer:- 38.2 g.

Solution:- The equation used for solving this type of calorimetry problems is:

q=mc\Delta T

where, q is the heat energy, m is mass, c is specific heat and delta T is the change in temperature.

Water temperature is increasing from 14.5 degree C to 50.0 degree C.

\Delta T=50.0-14.5  = 35.5 degree C

q is given as 5680 J and specific heat value is 4.186\frac{J}{g.^0C} .

The equation could be rearranged for m as:

m=\frac{q}{c*\Delta T}

Let's plug in the values in it:

m=\frac{5680}{4.186*35.5}

m = 38.2 g

So, the mass of water in the kettle is 38.2 g.


evablogger [386]3 years ago
3 0

Answer: The mass of the water in the kettle is 38.2 g

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Answer:

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mass of Glucose = 5 g

Mass of H₂O = ?

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                   6CO₂ + 6H₂O ----> C₆H₁₂O₆ + 6O₂

Solution:

First we have to find mass of glucose from balanced reaction.

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                        6CO₂ + 6H₂O -------> C₆H₁₂O₆ + 6O₂

                                     6 mol               1 mol

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Now

             6CO₂      +  6H₂O          --------->     C₆H₁₂O₆   +    6O₂

                              6 mol (18 g/mol)           1 mol (180 g/mol)

                                  108 g                            180 g

108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆)

So

if 108 g of water (H₂O) produce 180 g of glucose (C₆H₁₂O₆) so how many grams of water (H₂O) will be required to produce 5 g of glucose (C₆H₁₂O₆).

Apply Unity Formula

               108 g of water (H₂O) ≅ 180 g of glucose (C₆H₁₂O₆)

                X g of water (H₂O) ≅ 5 g of glucose (C₆H₁₂O₆)

Do cross multiply

                     mass of water (H₂O) = 108 g x 5 g / 180 g

                     mass of water (H₂O) = 3 g

So 3 g of water is required to produce 5 g of glucose.  

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