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Zinaida [17]
3 years ago
8

Assuming standard pressure, what physical constants are needed to calculate the change in enthalpy for raising a sample of water

from 250 K to 300 K. I) the heat capacity of ice II) the enthalpy of fusion of water III) the heat capacity of water IV) the enthalpy of vaporization of water V) the heat capacity of steam
Physics
1 answer:
Alik [6]3 years ago
8 0

Answer:

I) Heat capacity of ice (Sensible, ice), II) Enthalpy of fusion of water (Latent, fusion) and III) Heat capacity of water (Sensible, water).

Explanation:

At standard pressure, water has a fusion point and boiling point of 273.15 K and 373.15 K, respectively. Then, the total change in enthalpy is the sum of two sensible indicators and one latent indicator. That is:

\Delta H_{total} = \Delta H_{sensible, ice} + \Delta H_{latent, fusion} + \Delta H_{sensible, water}

The needed physical constants are: Heat capacity of ice (Sensible, ice), Enthalpy of fusion of water (Latent, fusion) and Heat capacity of water (Sensible, water).

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Answer: moving object :)

Explanation:

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What is the average velocity of a car that travels 30 kilometers due wet in 0.50
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60km/hr west

Explanation:

When you are dealing with velocity you always name the direction its going in

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3 years ago
To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 24.7 °C. You
drek231 [11]

Answer : The heat change of the cold water in Joules is, 1.6\times 10^3J

Explanation :

First we have to calculate the mass of cold water.

As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.

Density=\frac{Mass}{Volume}

Mass=Density\times Volume=1g/mL\times 45mL=45g

Now we have to calculate the heat change of cold water.

Formula used :

Q=m\times c\times (T_2-T_1)

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = 4.184J/g^oC

T_1 = initial temperature of cold water = 24.7^oC

T_2 = final temperature  = 33.4^oC

Now put all the given value in the above formula, we get:

Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC

Q=1638.036J=1.6\times 10^3J

Therefore, the heat change of cold water is 1.6\times 10^3J

4 0
3 years ago
Three metal spoons are on a table. They are each at room temperature. If the three spoons touch.
hoa [83]
If the three spoon touch nothing happens because they are all at room Temperature
3 0
3 years ago
Read 2 more answers
2. A student drives 7.8-km trip to school and averages a speed of
Alekssandra [29.7K]

Answer:

<em>The total time is: t=451.22 sec</em>

<em>The average speed is: V=34.57 m/s</em>

Explanation:

<u>Average speed</u>

The average speed is calculated by dividing the total distance traveled by an object (x) by the total time it took it to travel that distance (t).

\displaystyle V=\frac{x}{t}

Since the student makes the trip in two parts, we have to calculate the total distance and the total time.

We know the distance to school is 7.8 Km = 7,800 m. The student makes his way home over the same distance, thus the total distance is

x=2*7,800 m=15,600 m

The first trip to school was done at an average speed of v1=32.6 m/s. Knowing the distance and speed, we can calculate the time:

\displaystyle t1=\frac{x1}{v1}=\frac{7,800}{32.6}=239.26\ sec

The second trip back home was done at an average speed of v2=36.8 m/s. Let's calculate the second time:

\displaystyle t2=\frac{x2}{v2}=\frac{7,800}{36.8}=211.96\ sec

The total time is:

t=239.26\ sec+211.96\ sec=451.22\ sec

\boxed{t=451.22\ sec}

The average speed is:

\displaystyle V=\frac{15,600}{451.22}=34.57\ m/s

\boxed{\displaystyle V=34.57\ m/s}

6 0
3 years ago
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