# Specific heat capacity in physics

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#1
Hi I don’t know how to do this question. Can someone help? Thank you!

400g of oil at 80°C is mixed with 1.5kg of water. The mixture reaches equilibrium at 30°C. What was the initial temperature of the water?

The specific heat capacity of water is 4200 J/kg°C

The specific heat capacity of oil is 2100 J/kg°C
0
5 months ago
#2
Hey! A couple things to keep in mind:

Q=mc𐤃T (final temp in K - initial temp in K)
and
𐤃H=-(Q/n)

the mass of the oil burned would be substituted to find the number of moles (n) and the mass of water is substituted in the Q=mc𐤃T equation as it's the mass of the water that is heated.

Question: do you have access to a data booklet? If you do this question becomes a lot easier. Also, are you sure this isn't a chem question?
1
5 months ago
#3
(Original post by Sasuto)
Hi I don’t know how to do this question. Can someone help? Thank you!

400g of oil at 80°C is mixed with 1.5kg of water. The mixture reaches equilibrium at 30°C. What was the initial temperature of the water?

The specific heat capacity of water is 4200 J/kg°C

The specific heat capacity of oil is 2100 J/kg°C
Describe what's physically going to happen here in words, and how its going to happen (in terms of energy moving around-where does the heat to warm the water come from, does the oil cool in the mix, etc.) from that we can guide you to some equations. (the one you'll need is the one smc11 posted)

(Original post by Sm.c11)
Hey! A couple things to keep in mind:

Q=mc𐤃T (final temp in K - initial temp in K)
and
𐤃H=-(Q/n)

the mass of the oil burned would be substituted to find the number of moles (n) and the mass of water is substituted in the Q=mc𐤃T equation as it's the mass of the water that is heated.

Question: do you have access to a data booklet? If you do this question becomes a lot easier. Also, are you sure this isn't a chem question?
The oil is mixed and not burned (unless I'm misreading) - usually a common problem in phys though I've never seen it as 'oil mixed with water' o.o
2
5 months ago
#4
Where Ti is initial temperature of water, 1.5 x 4200 x (30 - Ti) = 0.4 x 2100 x (80 - 30).

Rearrange and solve to find Ti.

Important to note that the water is at a lower temperature so energy is transferred from the oil to water, which is why we do 30 - Ti as water's temperature is increasing and oil's temperature decreasing.
1
#5
(Original post by Sm.c11)
Hey! A couple things to keep in mind:

Q=mc𐤃T (final temp in K - initial temp in K)
and
𐤃H=-(Q/n)

the mass of the oil burned would be substituted to find the number of moles (n) and the mass of water is substituted in the Q=mc𐤃T equation as it's the mass of the water that is heated.

Question: do you have access to a data booklet? If you do this question becomes a lot easier. Also, are you sure this isn't a chem question?
Hi, thank you for your response! I was set this is physics! I’m quite new to the topic so I only know the Q=mc delta T equation. I don’t really understand the other equation 😅 I don’t think I’ve done it before. Thanks for the help though! Appreciate it a lot!
0
#6
Where Ti is initial temperature of water, 1.5 x 4200 x (30 - Ti) = 0.4 x 2100 x (80 - 30).

Rearrange and solve to find Ti.

Important to note that the water is at a lower temperature so energy is transferred from the oil to water, which is why we do 30 - Ti as water's temperature is increasing and oil's temperature decreasing.
Hi, thanks for the reply! I rearranged it and solved it and I got 23.3 as the initial temperature. But I’m not sure if it’s -23.3 or 23.3? Thank you!
0
5 months ago
#7
(Original post by Sasuto)
Hi, thanks for the reply! I rearranged it and solved it and I got 23.3 as the initial temperature. But I’m not sure if it’s -23.3 or 23.3? Thank you!
23.3, why would it be -23.3?
0
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