Please i cannot find any help on this question so far

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The discussion revolves around a heat transfer problem involving a rod with one end at 0 °C and the other at 100 °C. The user has calculated the heat flow at the cooler end to be 13.3 W but is struggling with determining the temperature gradient at the hot end given a temperature increase of 0.250 °C/sec. They attempt to apply the heat transfer equation but feel their approach is incorrect. The user seeks clarification on how to properly relate the heat flow and temperature changes to find the missing temperature gradient. The thread highlights the challenges of applying thermal conductivity principles in practical scenarios.
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please... i cannot find any help on this question so far!

Homework Statement


A rod is initially at a uniform temperature of 0 ^deg C throughout. One end is kept at 0 ^deg C, and the other is brought into contact with a steam bath at 100 ^deg C. The surface of the rod is insulated so that heat can flow only lengthwise along the rod. The cross-sectional area of the rod is 2.50 cm^2, its length is 120 cm, its thermal conductivity is 380 W/m * K, its density is 1.00 x 10^4 kg/m^3, and its specific heat capacity is 520 J/ kg * K. Consider a short cylindrical element of the rod 1.00 cm in length.

1. If the temperature gradient at the cooler end of this element is 140 deg C/m, how many joules of heat energy flow across this end per second?

2. If the average temperature of the element is increasing at the rate of 0.250 deg C/sec, what is the temperature gradient at the other end of the element?


Homework Equations





The Attempt at a Solution


I have gotten number 1 to be 13.3W, but do not know how to proceed with number 2. Any help would be greatly appreciated!
 
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both 1 & 2 are about a short cylindrical element of the rod 1.00 cm in length.

You know the heat going out at the cooler end. You know how quickly the temperature rises and you can calculate how much the internal energy of the element increases, so you should be able to calculate how much heat energy goes in at the hot end.
 
I am still stuggling. I see that from the equation H = kA(Th-Tc)/L, i can say that H is equal to dQ/dt, which is the rate of the heat current. I took this information and applied my .25 celsius degrees to the equation.

.25/(kA) = (Th-Tc)/L ---the temp. gradient

This is wrong though, what am I missing. I am desperate at this point!
 
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