Question

A technician measures the specific heat capacity of an unidentified liquid by immersing an electrical resistor...

A technician measures the specific heat capacity of an unidentified liquid by immersing an electrical resistor in it. Electrical energy is converted to heat, which is then transferred to the liquid for 124s at a constant rate of 65.3 W. The mass of the liquid is .779 kg and it’s temperature increases from 18.62 C to 22.51 C. A) find the average specific heat of the liquid in this temperature range. Assume that negligible heat is transferred to the container that holds the liquid and that no heat is lost to the surroundings.
Express your answer in joules per kilogram kelvin to three significant figures.
B) suppose that in this experiment heat transfer from the liquid to the container or its surroundings cannot be ignored. Is the result calculated in part (a) an overestimate or an underestimate of the average specific heat capacity? Explain

Homework Answers

Answer #1

Hello!

Here is the solution to your question!

Amount of heat as input = 65.3 W 124 s = 8097.2 J

Q = mc

so, c = Q/m

= 7800 J / (0.78 kg )

= 2506 J/kg/c

if some of the energy was used to heat the container , or the surroundings, then less energy would go into heating the liquid; if less energy caused the 3.99C temp increase, the specific heat must be less, so the value we have calculated is an overestimate.

For you to better understand this, let us sum up once again,

Energy supplied = 120*65 Ws = 8097 joules.
Energy supplied per kg = 7800/0.78 = 10000 joules/kg
rise in temperature = 22.54 - 18.55 C = 3.99C

10000 joules raised 1 kg by 3.99C

so 2506 joules raises 1kg by 1C


specific heat = 2506.2 joules perkgC

or can be written as 2.506 joules per gC

Hope this answer helps please eave a rating :)

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT