Composite Rod Question

  • Thread starter cwill53
  • Start date
  • #1
194
34
Homework Statement:
Two rods, one made of brass and the other made of copper, are joined end to end. The length of the brass section is 0.300m and the length of the copper section is 0.800 m. Each segment has cross-sectional area ##0.00500m^2##. The free end of the brass segment is in boiling water and the free end of the copper segment is in an ice–water mixture, in both cases under normal atmospheric pressure. The sides of the rods are insulated so there is no heat loss to the surroundings. (a) What is the temperature of the point where the brass and copper segments are joined? (b) What mass of ice is melted in 5.00 min by the heat conducted by the composite rod?
Relevant Equations:
$$H=\frac{dQ}{dt}=kA\frac{T_{H}-T_{C}}{L}$$
##k_{brass}=109\frac{W}{m\cdot K}## ##k_{copper}=385\frac{W}{m\cdot K}##
My answer:

a.
$$H=\frac{dQ}{dt}$$
$$H_{brass}=H_{copper}$$
$$(109\frac{W}{m\cdot K})(0.005m^2)\frac{100^{\circ}C-T_{c}}{0.3m}=(385\frac{W}{m\cdot K})(0.005m^2)\frac{T_{H}-0^{\circ}C}{0.8m}$$
$$T_{H,Cu}=T_{C,brass}=T_{2}$$
$$-1.8166667T_{2}+181.666667^{\circ}C=2.40625T_{2}\Rightarrow T_{2}=43.02^{\circ}C$$

b.
$$\frac{dQ}{dt}=103.5168J/s\Rightarrow \delta Q=(103.5168J/s)(5 min)=31055.04J$$
##c_{ice}=2100J/(kg\cdot C^{\circ})## ##L_{f,water}=334\cdot 10^3J/kg##

$$Q_{ice}=m(2100J/(kg\cdot C^{\circ}))(42^{\circ}C)+m(334\cdot 10^3J/kg)$$

$$31055.04J=(424342J/kg)m\Rightarrow m=0.073kg$$


Is this correct?
 

Answers and Replies

  • #2
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
2020 Award
36,210
6,825
I agree with your answer to a, but what relevance has 42C to part b?
 
  • #3
194
34
I agree with your answer to a, but what relevance has 42C to part b?
I just plugged the 42C in for Tc in the right hand side or Th in the left hand side to get dQ/dt....I don’t know if this is correct.
Should I have used 42C in finding the quantity of heat for ice?
 
  • #4
21,159
4,673
No. The ice is in an ice-water bath at 0 C, and the only thing happening is that the ice at 0 C is melting to form water at 0 C. It is not heating up to 42 C. The only place where the temperature is 42 C is at the junction between the copper and the brass.
 
  • #5
194
34
No. The ice is in an ice-water bath at 0 C, and the only thing happening is that the ice at 0 C is melting to form water at 0 C. It is not heating up to 42 C. The only place where the temperature is 42 C is at the junction between the copper and the brass.
Thanks for that. That alone just cleared up a lot of confusion I had. That means I should only be considering the heat produced from the ice changing phase heat lost to the ice-water bath.

From that I got m= 0.093 kg= 93 g.
 
Last edited:
  • #6
21,159
4,673
Thanks for that. That alone just cleared up a lot of confusion I had. That means I should only be considering the heat produced from the ice changing phase.

From that I got m= 0.093 kg.
I wouldn't call it heat produced. I would call it heat lost to the ice water bath. I haven't checked your math, but your approach is correct. Also, I would express the result as 93 grams.
 

Related Threads on Composite Rod Question

  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
7K
  • Last Post
Replies
1
Views
1K
Replies
1
Views
2K
  • Last Post
Replies
5
Views
931
  • Last Post
Replies
1
Views
3K
Replies
12
Views
548
  • Last Post
Replies
4
Views
6K
  • Last Post
Replies
1
Views
1K
Top