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3 years ago

13

Equal masses of substance A and substance B are at 25°C. Substance A has a lower specific heat capacity than substance B. If 50

cal of energy is transferred into each substance, which substance has a higher final temperature?

1 answer:

liraira [26]3 years ago

6 0

Dang bro that stuff is really hard I’m defiantly on a lower grade lol

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Treatment of (S)-( )-5-methyl-2-cyclohexenone with lithium dimethylcuprate gives, after protonolysis, a good yield of a mixture

tekilochka [14]

Answer:

use google and use the first link

Explanation:

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3 years ago

Charles is given two electrical conductors – aluminium and graphite. Help him to select one for making an electric wire. Justify

Iteru [2.4K]

Answer: aluminum I think

Explanation:

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3 years ago

A 5.0 L sample of gas at 300. K is heated to 600. K. What will the new volume of the gas be?

Ainat [17]

Answer:

$V_2=10L$

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the required new volume by using the Charles' law as a directly proportional relationship between temperature and volume:

$\frac{V_2}{T_2} =\frac{V_1}{T_1}$

In such a way, we solve for V2 and plug in V1, T1 and T2 to obtain:

$V_2=\frac{V_1T_2}{T_1}\\\\V_2=\frac{5.0L*600K}{300K}\\\\V_2=10L$

Regards!

4 0

2 years ago

How much energy (in Joules) is required to convert 129 grams of ice at −23.0 °C to liquid water at 18.0 °C?

Karo-lina-s [1.5K]

Answer:

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = <u>58902.9 joules</u>

Explanation:

This is a calorimetry problem:

Q = m . C . ΔT

Q = heat; m = mas; C is the specific heat and

ΔT = Final T° - Initial T°

Q = C lat . m

Q = Heat

m = mass

C lar = Latent heat of fusion

First of all we calculate the heat for ice, before it takes the melting point. (from -23°C to 0°C)

Q = 129 g . 2.10 J/g°C . (0°C - (-23°C)

Q = 129 g . 2.10 J/g°C . 23°C → 6230.7 joules

Then, the ice has melted. To be melted and change the state it required:

Q = C lat . m

Q = 333 J/°C . 129 g → 42957 joules

And in the end, we have water that changed its T° from O°C to 18°C

Q = 129 g . 4.184 J/g °C . (18°C - 0°C)

Q = 9715.2 Joules

The energy that is required for the process is:

6230.7 J + 42957 J + 9715.2 J = 58902.9 joules

5 0

3 years ago

You are burning wood to heat water for your industrial process. What is the mass of wood required to raise the temperature of 10

natta225 [31]

Answer:

18,8kg of wood

Explanation:

The energy you need to to raise the temperature of 1000 kg of water from 25.0 to 100.0 °C is:

q = C×m×ΔT

Where: q is heat, C is specific heat of water (4,184J/g°C), m is mass in grams (1000x10³g), and ΔT is 100,0°C - 25,0°C = 75,0°C

Replacing:

q = 4,184J/g°C×1000x10³g×75,0°C

<u><em>q = 3,14x10⁸ J of heat are required</em></u>

<u><em /></u>

Now, if the heating value of dry wood is 16,72 MJ/kg = 16,72x10⁶ J/kg, mass of wood required is:

3,14x10⁸J × (1kg / 16,72x10⁶ J) = <em>18,8 kg of wood are required</em>

<em></em>

I hope it helps!

5 0

3 years ago