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I'm cooking frozen peas in my kitchen. Experiment 1 - I take a quantity of frozen peas, put them into a (room temperature) saucepan, add a quantity (at least enough to cover the peas) of boiling water from the kettle and immediately place the saucepan on the gas hob. Experiment 2 - I then repeat the process, the only difference being I increase the quantity of boiling water added to the peas. Will there be any difference in the two experiments between the time the water takes to come to the boil? Assume the saucepan is big enough to comfortably contain the water and peas.

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Let $m_1$ be the mass of the peas, which we'll assume to be mainly water, $m_2$ the mass of boiling water added and $m_p$ the mass of the pan. We'll assume no heat losses.

After adding everything to the pan and allowing for thermal equilibrium to be established the temperature of pan, water and peas is $T$.

$T$ can be calculated from the adiabatic heat balance, before/after equilibrium is achieved:

$$m_p C_{p} (T-T_0) + m_1 C_w (T-0)+ Lm_1+m_2 C_w (T-100)=0,$$

where $C_{p}$, $C_{w}$ are the respective specific heat capacities of the pan and the water (or peas) and $T_0$ the initial temperature of the pan and $L$ the specific latent heat of melting of the peas.

$T$ can be isolated as:

$$T=\frac{m_p C_pT_0+100m_2C_w-Lm_1}{m_pC_p+m_1C_w+m_2C_w}.$$

We can see that, as we increase $m_2$ then $T$ increases (all other things being equal), as can be expected.

The question now becomes, how much more heat do we have to add to get to $100^{\circ}$? That can be calculated as follows:

$$\Delta Q=m_pC_p(100-T)+m_1C_w(100-T)+m_2C_w(100-T).$$

Or:

$$\Delta Q=(m_pC_p+m_1C_w+m_2C_w)(100-T).$$

Or:

$$\Delta Q=100(m_pC_p+m_1C_w+m_2C_w)-T(m_pC_p+m_1C_w+m_2C_w).$$

Using the expression for $T$, we can then write:

$$\Delta Q=m_pC_p(100-T_0)+100C_wm_1+Lm_1.$$

Note that this expression does not contain $m_2$, so $\Delta Q$ is in fact invariant to the mass of boiling water initially added. The final expression for $\Delta Q$ is actually the amount of heat needed to thaw the peas and heat peas and pan to $100^\circ$.

Adding more boiling water to the frozen peas does not cut down on heating time.

However, adding more boiling water does cost more energy, as the Enthalpy of the boiling water also has to be accounted for. The fastest and most economical way of cooking frozen peas is thus not to use any boiling water at all.

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  • $\begingroup$ Thanks. I thought that would be the case but didn't have the physics to back it up. $\endgroup$ – Peter4075 Nov 10 '15 at 11:21
  • $\begingroup$ @Peter4075: thanks for the upvote. I'm going to make a small edit now. $\endgroup$ – Gert Nov 10 '15 at 16:00
  • $\begingroup$ Must admit, I've never heard of cooking frozen peas without water. It works - I've just tried it - but you need to be careful, the saucepan gets very hot (my stainless steel saucepan began to discolour with the heat). Don't burn down the kitchen! $\endgroup$ – Peter4075 Nov 10 '15 at 19:39
  • $\begingroup$ @Peter4075: I always use a microwave oven for this, with a smattering of butter, no water. Works always! $\endgroup$ – Gert Nov 10 '15 at 19:41
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Will there be any difference in the two experiments between the time the water takes to come to the boil?

No, if we assume no heat is radiated in the environment.

As long as the water is at 100 C and you also want the mass of peas to rise its temperature from 0 to 100 C you can leave the water as it is, without heating it any more, and you heat just the peas which require a clear quantity of energy, $Q$, to rise their temperature to 100 C. So, no matter how much water at 100 C you have the quantity of energy required by the mass of peas to reach 100 C will be the same. As the flame deliver a known quantity of heat per second, q, and Q = q*t, it means that t = Q/q = constant and independent of the mass of water at 100 C.

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