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zhannawk [14.2K]
2 years ago
13

Beth heats a pot of water to cook pasta. What may happen as a result of adding heat?

Chemistry
2 answers:
Ivan2 years ago
8 0

Answer: i’m taking a test w this question rn and i think the best answer is D. When the water begins to boil, some of the water molecules will gain enough energy to change into a gas. i could def be wrong tho

n200080 [17]2 years ago
6 0
The water will started boiling and the pasta is going to start like melting so it can get a soft and smooth texture so you can eat it.
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In an electrically heated boiler, water is boiled at 140°C by a 90 cm long, 8 mm diameter horizontal heating element immersed in
RideAnS [48]

Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

4 0
3 years ago
100.0 g of 4.0°C water is heated until its temperature is 37.0°C. If the specific heat of water is 4.184 J/g°C, calculate the am
Lady bird [3.3K]

Answer:

13.8072 kj

Explanation:

Given data:

Mass of water = 100.0 g

Initial temperature = 4.0 °C

Final temperature = 37.0°C

Specific heat capacity = 4.184 j/g.°C

Heat absorbed = ?

Solution:

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = 37.0°C -  4.0 °C

ΔT = 33.0°C

Q = 100.0 g ×4.184 j/g.°C × 33.0°C

Q = 13807.2 j

Joule to KJ:

13807.2 j  × 1kj  /1000 j

13.8072 kj

5 0
3 years ago
Please help me with this chemistry
Elena-2011 [213]

Answer: gasoline, water, sea water, chloroform and mercury so B

Explanation:

6 0
2 years ago
Using the volumes of sodium thiosulfate solution you just entered, the mass of bleach sample, and the average molarity of the so
dedylja [7]

Answer:

3.18 (w/w) %

Explanation:

In the problem, you can find mass of NaClO knowing the reaction of NaClO with Na₂S₂O₃ is:

NaClO + 2Na₂S₂O₃ + H₂O → NaCl + Na₂S₄O₆ +2NaOH + NaCl

<em>Where 1 mole of NaClO reacts with 2 moles of Na₂S₂O₃</em>

<em> </em>Moles of thiosulfate in the titration are:

0.0101L ₓ (0.042mol / L) = 4.242x10⁻⁴ moles of Na₂S₂O₃

Thus, moles of NaClO in the initial solution are:

4.242x10⁻⁴ moles of Na₂S₂O₃ ₓ (1mol NaClO / 2 mol Na₂S₂O₃) = 2.121x10⁻⁴ moles NaClO

As molar mass of NaClO is 74.44g/mol, mass of 2.121x10⁻⁴ moles are:

2.121x10⁻⁴ moles ₓ (74.44g / mol) = <em>0.0158g of NaClO</em>

As mass of bleach is 0.496g, mass percent is:

0.0158g NaClO / 0.496g bleach ₓ 100 =

<h3>3.18 (w/w) % </h3>
4 0
3 years ago
I need help with this!!
8_murik_8 [283]

Answer:

0.84kg of gatorade powder is needed.

Explanation:

From the question given above, we were told that the team manager mixes 0.6kg of gatorade powder with 5 gallons of water.

Now, to obtain the desired amount of the powder of gatorade needed for 7 gallons of water, we simply do the following:

If 0.6kg of gatorade powder required 5 gallons of water,

Then Xg of gatorade will require 7 gallons of water i.e

Xg of gatorade = (0.6 x 7)/5

Xg of gatorade = 0.84kg

Therefore, to maintain the same gatorade mixture, the team manager must mix 0.84kg of gatorade powder with the 7 gallons of water.

4 0
3 years ago
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