- #1
cwill53
- 220
- 40
- 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?
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?
