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Q : An electric heater of power $1000 \mathrm{~W}$ raises the temperature of $5 \mathrm{~kg}$ of a liquid from $25^{\circ} \mathrm{C}$ to $31^{\circ} \mathrm{C}$ in 2 minutes. Calculate : (i) the heat capacity and (ii) the specific heat capacity of liquid.

Solution:

Time $t=2$ minutes $=2 \times 60 \mathrm{~s}=120 \mathrm{~s}$ Energy supplied by the heater $=$ power $\times$ time $\Delta Q=1000 \mathrm{~W} \times 120 \mathrm{~s}=1 \cdot 2 \times 10^{5} \mathrm{~J}$ Energy used in raising the temperature of liquid from $25^{\circ} \mathrm{C}$ to $31^{\circ} \mathrm{C}$

  • $\Delta Q^{\prime}=$ heat capacity $\times$ rise in temperature $=C^{\prime} \times(31-25)=6 C^{\prime} \mathrm{J}$
  • If there is no loss of heat energy ,

Energy supplied $\Delta Q=$ Energy used $\Delta Q^{\prime}$ $\therefore \quad 1.2 \times 10^{5}=6 C^{\prime}$ Heat capacity $C^{\prime}=\frac{1 \cdot 2 \times 10^{5}}{6}=2 \times 10^{4} \mathbf{J} \mathbf{~}^{-1}$

My Q is the point where we assume no assume no loss heat energy.

How can we determine I.e what will be the method for finding the amount of loss of heat energy. Let us say if there is any.

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  • $\begingroup$ there will always be assumptions. think about what makes an assumption reasonable. why would it be reasonable to assume another sunrise sometime in the next 25 hours? $\endgroup$
    – Abel
    Commented Nov 30, 2021 at 12:32
  • $\begingroup$ @Abel Because of the way the sun,earth & moon rotate & revolve around makes it reasonable. $\endgroup$
    – S.M.T
    Commented Nov 30, 2021 at 13:39
  • $\begingroup$ We need more data on the problem normally, for example the heat conductivity with the atmosphere. Then we can model the problem using more equations.e.g the heat equation. In this case, the system is fully defined i.e all the heat is already used to raise the temperature. $\endgroup$
    – luis_
    Commented Nov 30, 2021 at 14:36
  • $\begingroup$ This is the only assumption you can/need to make because there is no data for you to calculate the loss. If you don't feel comfortable, you can make the assumption more specific by saying "Under ideal condition, assume... $\endgroup$
    – r13
    Commented Nov 30, 2021 at 20:39
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    $\begingroup$ This is perhaps the only case in real life engineering where the conversion efficiency actually is 100%. Electrical resistive heating, or Joule Heating, is 100% efficient, because it outputs 0% work and 100% "waste heat" which in this case is what you are after. The only "losses" in this process (say in a water heater) would be heat that escaped through the walls and did not heat the water. $\endgroup$
    – RC_23
    Commented Dec 1, 2021 at 5:12

2 Answers 2

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Don't make it more complicated than it needs to be.

If they were asking a heat transfer problem they would include the data required - ambient temp, materials, insulation, liquid properties for convective transfer, etc. This is just a problem for heat properties of the liquid. The assumption is zero heat loss from the system.

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Right now, you are given power and time, which allows you to work out $Q_{electric}$.

$$Q_{electric} = P \ t$$

In the real world, you would require the conversion efficiency, but you are not given that. Due to losses actual heat required to heat something must be less than electric heat produced ($Q < Q_{electric}$).

$$Efficiency = \eta = \frac{Q} {Q_{electric}} \times 100\% $$

Assuming efficiency is 100%.

$$Q = \eta\ Q_{electric} = 100\% \times Q_{electric} = Q_{electric}$$

$$Q = M\ C\ ΔT$$

This will allow you to work out specific heat capacity of liquid.

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