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adelina 88 [10]
3 years ago
12

How much heat is required to convert 0.3 kilogram of ice at 0°C to water at the same temperature

Physics
2 answers:
Grace [21]3 years ago
6 0

Answer:

100,200J of heat is required to convert 0.3kg of ice of 0°C to water at same temperature.

Explanation:

Heat = mass * lf

Latent heat of fusion (lf) of water is 334J/g

Heat = 300g * 334 J/g

Heat = 100,200J of heat

irina1246 [14]3 years ago
6 0

Answer:

100800 J

Explanation:

Specific Latent heat of fusion of water: This is the quantity of heat required to change the unit mass of water from ice to liquid without change in temperature.

The expression for Specific latent heat of fusion is given as

Q = Cm...................... Equation 1

Where Q = Amount of heat, C = Specific latent heat of fusion of ice, m = mass of ice.

Given: m = 0.3 kg,

Constant: C = 336000 J/kg

substitute into equation 1

Q = 336000(0.3)

Q = 100800 J.

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Answer:

The length of the object would shrink to zero which is not possible.

Explanation:

A rocket or any body cannot reach the speed of light because according to theory of relativity the and the Lorentz factor the length of the object would shrink to zero and the time dilation for that body would be infinite.

The Lorentz factor is given as:

\gamma=\frac{1}{\sqrt{\frac{v^2}{c^2} } }

where:

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c = speed of light

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A car travels a distance of 100 km. For the first 30 minutes it is driven at a constant speed of 80 km/hr. The motor begins to v
gregori [183]

Explanation:

First, we need to determine the distance traveled by the car in the first 30 minutes, d_{\frac{1}{2}}.

Notice that the unit measurement for speed, in this case, is km/hr. Thus, a unit conversion of from minutes into hours is required before proceeding with the calculation, as shown below

                                          d_{\frac{1}{2}\text{h}} \ = \ \text{speed} \ \times \ \text{time taken} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ \left(\displaystyle\frac{30}{60} \ \text{h}\right) \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 80 \ \text{km h}^{-1} \ \times \ 0.5 \ \text{h} \\ \\ \\ d_{\frac{1}{2}\text{h}} \ = \ 40 \ \text{km}

Now, it is known that the car traveled 40 km for the first 30 minutes. Hence, the remaining distance, d_{\text{remain}} , in which the driver reduces the speed to 40km/hr is

                                             d_{\text{remain}} \ = \ 100 \ \text{km} \ - \ 40 \ \text{km} \\ \\ \\ d_{\text{remain}} \ = \ 60 \ \text{km}.

Subsequently, we would also like to know the time taken for the car to reach its destination, denoted by  t_{\text{remian}}.

                                              t_{\text{remain}} \ = \ \displaystyle\frac{\text{distance}}{\text{speed}} \\ \\ \\ t_{\text{remain}} \ = \ \displaystyle\frac{60 \ \text{km}}{40 \ \text{km hr}^{-1}} \\ \\ \\ t_{\text{remain}} \ = \ 1.5 \ \text{hours}.

Finally, with all the required values at hand, the average speed of the car for the entire trip is calculated as the ratio of the change in distance over the change in time.

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Therefore, the average speed of the car is 50 km/hr.

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<em>A. A positive and negative charge attract each other.</em>

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