1one correction in varun's answer
after connection with rod the rod not only loose heat by conduction but also through radiation
so
\frac{dQ}{dT}= [\frac{dQ}{dT}]_{cond.} + [\frac{dQ}{dT}]_{radiation}
A solid body X of heat capacity C is kept in a an atmosphere whose temperature is TA = 300 K. At time t = 0, the temperature of X is T0 = 400 K. It cools according to Newton’s law of cooling. At time t1 its temperature is found to be 350 K.
At this time (t1) the body X is connected to a large body Y at atmospheric temperature TA through a conducting rod of length L, cross-sectional area A and thermal conductivity K. The heat capacity of Y is so large that any variation in its temperature may be neglected. The cross-sectional area A of the connecting rod is small compared to the surface area of X. Find the temperature of X at time t = 3t1.
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9 Answers
1357During the melting of a slab of ice at 273K at atmospheric pressure :
(A) positive work is done by the ice-water system on the atmosphere.
(B) positive work is done on the ice-water system by the atmosphere.
(C) the internal energy of the ice-water system increases.
(D) the internal energy of the ice-water system decreases.
- Ayush Lodha the answer will be d as the system is open and the internal energy of the system does some work against the atmospheric pressure and so the internal energy of the icewater system will decrease .
Upvote·0· Reply ·2013-02-10 02:38:18
1see manish i could only proceed only till this far
since in the first part ....... d\Theta /dt=-k(\Theta -300)
so we know the tme in propotion to k.... where k=k=e\alpha A/mc so we know it ....
and after
some time
dq/dt=\Delta T/R
and q=mcdletaT and r=l/ka...... so i think here \Delta t=(\Theta -300)
and we can solve it form here am i rite but personally i feel this sum is too advanced
11so decoder radiation will be propotional to newtons law of cooling
1357I forgot this question :
1.
initially the cooling will be due to newton's law of cooling only.
So we have
dQ/dt=CdT/dt=-h(T-300)
or dT/(T-300)=-h/C dt
or ln2=ht1/C or h/C=ln2/t1
In the next case, heat loss will be due to radiation as well as conduction:
dQ/dt=CdT/dt=-h(T-300)-KA(T-300)/L
dT/(T-300)=-[h/C+KA/CL]dt
ln[(T-300)/50]=-[ln2/t1+KA/L]2t1
ln[(T-300)/50]=-[2ln2+2KAt1/L]
(T-300)/50=e-[2ln2+2KAt1/L]=(1/4)e2KAt1/L
T=12.5e2KAt1/L+300