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Specific Heat Capacity

by David7812
Tags: capacity, heat, specific
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David7812
#1
Oct5-08, 02:34 PM
P: 3
I am stuck on my Physics work. I have already completed the first 10 questions but I do not understand these, can someone please help me with a method of how to find out.

1. When 0.25kg or water at 70C is mixed with water at 20C the final temperature of the mixture is 30C. find the mass of the cold water. Specific heat capacity of water -
4200J/KG/C.

2. 200g of copper (420J/KG/C) at 80C is put into 150g of oil (600J/KG/C). If the final temperature of the mixture is 50C find the starting temperature of the oil.

3. 400g of water at 20C is placed in a 200g Aluminium container (900J/KG/C) also at 20C. If 6000J are added find the final temperature of the oil and container.

4. 1.2kg of oil (600J/KG/C) at 150C is placed into a copper container (400J/KG/C) which mass is 2kg. If the starting temperature of the container is 25C find the final temperature of the oil and container.

5. How much heat energy is needed to raise the temperature of 1kg of ice at 0C to water at 40C. Specific Latent Heat of ice = 334KJ/KG,

6. 2kg of solid wax at its melting point 65C is placed into 3kg of oil at 150C, find the final temperature of the mixture. Specific heat Latent of wax = 4000J/KG, Specific Heat Capacity of liquid wax = 800J/KG/C, specific heat capacity of oil = 2500J/KG/C.
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dlgoff
#2
Oct5-08, 03:15 PM
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Welcome to PF. We need to see some work from you before helping. Have you read the forum rules?
David7812
#3
Oct5-08, 03:19 PM
P: 3
Some of my work:

500g of ice at -12C is heated to water at 5C. How much energy is required?

Heat = (0.5x2100x12)+(0.5x330000)+(0.5x4200x5)
Heat = 12600 + 165000 + 10500
Heat = 188100J

Kurdt
#4
Oct5-08, 03:30 PM
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Specific Heat Capacity

What does that have to do with any of the questions you've posted?
David7812
#5
Oct5-08, 03:32 PM
P: 3
It was part of my homework. Please can someone help me? I also have the common cold so I can't think straight.
pdpax
#6
Apr15-11, 05:47 AM
P: 3
I go straightly to the answers:

Let d=delta, t=temperature, Q=heat, c=Specific heat capacity, m=mass.

1)

Q=mcdt=0.25(4200)(30-70)=-42000
=> Q= mcdt= m(4200)(30-20) = 42000
=> m= 1 Kg.

2)

Q[lost by Copper]= -Q[taken by Oil]
=> mcdt=mcdt
=> 0.2(420)(50-80)=-0.15(600)(50-t)
=> t=22 C

3)

Q = mcdt + mcdt = 6000
=> dt[0.4(4200)+0.2(900)]=6000
=> dt= 6000/1860
=>t[final]=20+6000/1860 ≈ 23.2258

4)

t[final] = (mct + mct)/(mc+mc) = [1.2(600)(150) + 2(400)(25)] / [1.2(600) + 2(400)]
=> t[final] ≈ 84.21 C

5)

Q[total] = Q[L[f]] + Q = mL[f] + mcdt = 1(334000) + 1(4200)(40) = 502000 J = 502 KJ

6)

Q[taken by the Solid wax] = Q[L[f]] = mL[f] = 2(4000)=8000 J
Q[Given by the oil to the Solid wax] = mcdt = 3(2500)(t[final1]-150) = -8000
=> t[final1] ≈ 148.93 C

t[Final]= (mct+mct)/(mc+mc) = [(2(65)(800) + (3(2500)(150))]/[2(65)+3(2500)]
=> t[Final] ≈ 134.175 C
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