Specific Heat Capacity of Water - Physics Experiment

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To find the specific heat capacity of water, you need a power pack, a joulemeter, an electronic balance, and a calorimeter.
Place the calorimeter on the balance, zero the balance, and pour in some water. Take a reading of the mass of the water from the balance.
Plug the power pack into the joulemeter and plug the joulemeter into the calorimeter.
Place a thermometer in the calorimeter and note the initial temperature.
Turn on the power pack and allow the temperature of the water to increase. Once the temperature has been raised to a significant amount (I increased it by 7 degrees in the video), record the amount of energy transferred and the final temperature.
Calculate the specific heat capacity of water using the equation:
energy transferred = mass x SHC x temperature change

Опубликовано:

8 май 2021

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Комментарии : 32

@swaggyllama 2 года назад

This video helped more than my actual school

@ali3rady313 Год назад

كنت ابحث عن اجابة هذا السؤال من وقت طويل وها قد حصلت عليه من عندك اشكرك صديقي

@NoumanNajeebRana 2 года назад

great video, helped a lot with my students.

@alexisjones419 2 года назад

This has ASMR potential.

@vt.physics 2 года назад

:D So true!

@aaditjivani3905 Год назад

I really loved this video! Can I get the name and link of the calorimeter you are using so that I can use this for my Internal assessment experimentation?

@vt.physics Год назад

It's not exactly the same version, but one of these would work: edulab.com/product/calorimeter-set-copper-joules/

@ryleebrownfox 22 дня назад

there is something wrong with the STUDENT DIGITAL JOULEMETER. I use the formula Energy = Power x Time where Power = I x I x R ( I = 1amp, R = 6 Ohms) Power = 1amp x 1amp x 6 ohms Power = 6 Watts (J/Sec) Energy = (6 J/Sec ) x (498 Sec) Energy = 2988 Joules Then c = E / ( m x dT) c = 2988 J / (0.099kg x 7°C) c = 4311 J/kg.°C A lot closer to 4186 J/kg°C. I think that if we can measure the correct current consumption by using a multimeter, we could get the actual power or energy absorbed by the water since we have the resistance of 6 ohms to base it from. I'm a beginner so please correct me if I'm wrong.

@MManju-zg5bt 3 года назад

Super. Madam

@actiagodc Год назад

Very good. Could you inform the values of voltage, electric current and resistance used in the experiment? Thank you.

@vt.physics Год назад

I believe it was a 12 V power supply. I connected it directly to the immersion heater and didn’t use any external resistor

@actiagodc Год назад

@@vt.physics thank's.

@everythingtv4875 2 года назад

when do i stop the joulemeter? is it after it increases by 10° or after 10 minutes? pls answer

@vt.physics 2 года назад

You can switch off the power source and stop the joulemeter after you've seen a significant amount of temperature change. In my case, I've just waited a few minutes for temperature to raise by 7 degrees C. Time is not a factor in the specific heat capacity equation.

@MManju-zg5bt 3 года назад

U r videos are awesome madam

@vt.physics 3 года назад

Thank you so much 🙂

@MManju-zg5bt 3 года назад

Madam name please 😁

@MManju-zg5bt 3 года назад

I am from India

@dionisienatea3137 11 месяцев назад

hi. i was doing this experiment in a more controlled environment with high precision temperature measuring equipment and energy meters, vacuum insulated water container, 4.5kg of distilled water-to minimize the errors and every single time i got values for Cp around 4724 J/kgC. I run out of ideas of what could create the difference. Can we find out who and when and how were calculated values for water what we take for granted? i calculated the thermal losses to minute values and still doesn't get even close to the book values for water Cp-the difference is too high. Can anyone please give me an idea?

@xuanatngo4111 5 месяцев назад

Chào bạn, tôi đến từ Vietnam và tôi nghĩ vấn đề ở đây là bạn chưa thể đo được năng lượng nhiệt được truyền vào để làm nóng thiết bị của bạn. Ta không thể bỏ qua nhiệt lượng truyền vào thiết bị đun vì nó thật sự rất đáng kể.

@dionisienatea3137 5 месяцев назад

Thank you. I figured it out. I was using low heating power and took too long to have a reliable amount of data. I redone the experiment with high power heater and shorter time, to avoid external energy dissipation and i got values very close to the book.@@xuanatngo4111

@MManju-zg5bt 3 года назад

From India ....

@superrandomuser 2 года назад

stop

@nohaahmed3027 10 месяцев назад

What if i dont have jolmeter how can i calculate amount og energy

@vt.physics 10 месяцев назад

Do you have a voltmeter and an ammeter to measure voltage and current? If so, you can calculate the power. Then use the equation energy=power x time

@tonychau9560 Год назад

what would be the sources of error?

@vt.physics Год назад

Zero error from the electronic balance and parallax error from not reading the thermometer at eye level are examples of possible errors. Also, some energy would have been used to raise the temperature of the container