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Fofino [41]
3 years ago
9

How much heat is required to convert 500g of liquid water at 28°C into steam at 150°C? Take the specific heat capacity of water

to be 4183J/kg°C and the latent heat of vaporization to be 2.26×10⁶J/kg
Physics
1 answer:
JulijaS [17]3 years ago
8 0

Answer:

1,327,063Joules

Explanation:

Heat energy is the energy needed to convert the state of a body from one phase to another.

According to the question, we want to calculate the total heat required to convert water into vapour (steam).

Note that before water can vapourize, it has to reach the boiling point first which is at 100°C. Heat energy needed to convert the water to 100°C is expressed as H1 = mcΔθ

m is the mass of the object in kg =0.5kg

c is the spcific heat capacity of water = 4183J/kg°C

Δθ is the change in temperature = 100-28 = 72°C

H1 = 0.5*4183*72

H1 = 150,588Joules

Energy required to convert the water to team H2 =mLsteam

Lsteam is the latent heat of vaporization = 2.26×10⁶J/kg

H2 = 0.5*2.26×10⁶

H2 = 1130000Joules

Heat energy needed to convert the water to 150°C is expressed as H3 = mcΔθ

m is the mass of the object in kg =0.5kg

c is the spcific heat capacity of steam= 1859J/kg°C

Δθ is the change in temperature = 150-100 = 50°C

H3 = 0.5*1859*50

H1 = 46,475Joules

Total Heat requires = H1+H2+H3 = 150,588Joules+1130000Joules+ 46,475Joules = 1,327,063Joules

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Will mark as brainliest if correct!!!!!!!!!!!!!!!!!!!!!!!!!
Digiron [165]

Answer:

Incident ray

Explanation:

"To describe the reflection of light, we will use the following terminology. The incoming light ray is called the incident ray. The light ray moving away from the surface is the reflected ray. The most important characteristic of these rays is their angles in relation to the reflecting surface."

https://www.siyavula.com/read/science/grade-11/geometrical-optics/05-geometrical-optics-03

3 0
3 years ago
Read 2 more answers
A 75-g projectile traveling at 600 m /s strikes and becomes embedded in the 40-kg block, which is ini-tially stationary. Compute
levacccp [35]

Explanation:

The given data is as follows.

          Mass, m = 75 g

         Velocity, v = 600 m/s

As no external force is acting on the system in the horizontal line of motion. So, the equation will be as follows.

          m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

where,  m_{1} = mass of the projectile

            m_{2} = mass of block

              v = velocity after the impact

Now, putting the given values into the above formula as follows.

              m_{1}v_{1_{i}} = (m_{1} + m_{2})vi

         75(10^{-3}) \times 600 = [(75 \times 10^{-3}) + 50] \times v

                                  = \frac{45}{50.075}

                              v = 0.898 m/s

Now, equation for energy is as follows.

               E = \frac{1}{2}mv^{2}

                  = \frac{1}{2} \times (75 \times 10^{-3} + 50) \times (600)^{2}

                  = 13500 J

Now, energy after the impact will be as follows.

             E' = \frac{1}{2}[75 \times 10^{-3} + 50](0.9)^{2}

                 = 20.19 J

Therefore, energy lost will be calculated as follows.

           \Delta E = E  E'

                       = (13500 - 20) J

                       = 13480 J

And,   n = \frac{\Delta E}{E}

             = \frac{13480}{13500} \times 100

             = 99.85

             = 99.9%

Thus, we can conclude that percentage n of the original system energy E is 99.9%.

7 0
3 years ago
Read 2 more answers
A solid object has a mass of 104 kg and a volume of 1,278 m3. What is its density? 0.081 kg/m3 12.29 g/cm3 132912.00 g/cm3 canno
REY [17]
The density is 81.4 g/m3. Before you start plugging numbers into the density formula (D=M/V), you should convert 104 kg to grams, which ends up being 104,000 grams. Then you can plug in the 104,000 grams and 1,278 m3 into the formula. When you divide the mass by the volume, you get a really long decimal, which you can round to 81.4 g/m3, or whatever place your teacher wants you to round to.
4 0
3 years ago
Read 2 more answers
Which of the following would be considered a "point source" of water
KATRIN_1 [288]

Answer:

c. Groundwater contamination at a fracking site

Explanation:

all others could be sources from tens of square kilometers of surface area.

Fracking is limited to within a short range of the well hole.

5 0
3 years ago
Water is used to cool down automobile engines when they get hot. Why is water used as a coolant? A. Water is a good conductor. B
Mrac [35]

Water is usually used to cool down automobile engines when they get hot, yes. Therefore, that means water has a high heat capacity.

That makes the answer letter D you provided above.

D) Water has a high heat capacity.

Another example would be trying to put out a fire with a bucket of water. Usually, you can put out the fire debating on the size!

3 0
3 years ago
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