Physics refraction and Specific heat capacity problem

In summary, the first question is discussing the specific latent heat of fusion of ice, while the second question is discussing the formula for refractive index and the use of the Law of Sines.
  • #1
usman975
1
0
I am a IGCSE student!


Ok so in the physics w08 qp ( http://www.xtremepapers.net/CIE/Cambridge IGCSE/0625 - Physics/0625_w08_qp_03.pdf )

Qestion 5 c Using a 40 W heater, 16.3 g of ice is melted in 2.0 minutes. The heater is then switched
off. In a further 2.0 minutes, 2.1 g of ice is melted.
Calculate the value of the specific latent heat of fusion of ice from these results.

This means that the heater has melted 16.3g of Ice and the other 2.1 is irrelevant as it says that its been melted after the heater has been switiched off

BUT IN THE MARKING SCHEME ( http://www.xtremepapers.net/CIE/Cambridge IGCSE/0625 - Physics/0625_w08_ms_3.pdf )

(mass of ice melted by heater = 16.3 – 2.1) = 14.2 g C1
ml in any form, words, symbols or numbers C1
Wt or Pt in any form, words, symbols or numbers accept VIt C1
338 J/g OR 338 000 J/kg c.a.o A1

I DONT understand how the heater melted 14.2g?? its in the question that it melted 16.3 and the other was meleted after it was switiched off.




Now the other question same paper Q6 b) Check it from the paper

n = sinr/sini OR n = sini/sinr in any form C1
sinr/sin30 = 1.49 OR sinr = 1.49 × sin30 C1
48.0° – 48.2° A1

this is the answer. I don't understand that the formula for refractive index is SinI/Sinr
its stated in the answer above too. THEN HOW CAN IT GET Sinr=1.49 x Sin30
It should be sinr=Sin30/1.49


PLZZZ HELp
 
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  • #2
In the first question, the 16.3 g of ice that was melted by the heater is actually 14.2 g when you subtract the 2.1 g of ice that was melted after it was switched off. In the second question, the formula for refractive index is n = sin i/sin r. However, the answer is actually showing the reverse, where sin r/sin i = 1.49. This is because the question is asking for the angle of refraction, which requires the use of the Law of Sines, which states that sin i/sin r = n. Therefore, to find the angle of refraction, you need to solve for sin r by dividing both sides of the equation by sin i, giving the result of sin r = n × sin i. So in this case, sin r = 1.49 × sin 30.
 

Related to Physics refraction and Specific heat capacity problem

1. What is refraction in physics?

Refraction is the phenomenon where light waves change direction and speed as they pass through different mediums, such as air, water, or glass. This occurs because the speed of light is different in different substances, causing the light to bend.

2. How is the index of refraction calculated?

The index of refraction is calculated by dividing the speed of light in a vacuum by the speed of light in a specific medium. It is represented by the symbol "n" and is a unitless value.

3. What is Snell's Law?

Snell's Law is a formula that describes the relationship between the angle of incidence and angle of refraction for a light ray passing through two different mediums. It states that the ratio of the sine of the angle of incidence to the sine of the angle of refraction is equal to the ratio of the speed of light in the first medium to the speed of light in the second medium.

4. How does specific heat capacity affect the temperature of an object?

Specific heat capacity is the amount of heat energy required to raise the temperature of one unit of mass of a substance by one degree Celsius. The higher the specific heat capacity of a substance, the more heat energy is needed to raise its temperature. Therefore, substances with a higher specific heat capacity will have a slower rate of temperature change compared to substances with a lower specific heat capacity.

5. What factors affect the specific heat capacity of a substance?

The specific heat capacity of a substance is affected by its mass, temperature, and chemical makeup. Substances with a larger mass will require more heat energy to raise their temperature compared to substances with a smaller mass. Additionally, substances with strong intermolecular forces, such as water, tend to have a higher specific heat capacity than substances with weaker intermolecular forces, such as metals.

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