If liquefied oxygen at 1 atmospheric pressure | vedclass

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Question

$Q=m c \Delta T$

$\mathrm{Q}=\mathrm{mc}\left(\mathrm{T}-\mathrm{T}_{0}\right)\,\,\,\,\,\,\,...(i)$

$Q=$ Kt whereas $K$ is heating rate

from

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Vedclass · Answer

$Q=m c \Delta T$

$\mathrm{Q}=\mathrm{mc}\left(\mathrm{T}-\mathrm{T}_{0}\right)\,\,\,\,\,\,\,...(i)$

$Q=$ Kt whereas $K$ is heating rate

from $50$ to boiling temperature, $\mathrm{T}$ increases linearly.

At vaporization, equation is $Q=m L$

so, temperature remains constant till vaporisation is complete

After that, again Eqn $(i)$ is followed and temperature increases linearly

Column $-I$	Column $-II$
$(a)$ Combined existence of liquid-gaseous state of substance.	$(i)$ Sublimation curve
$(b)$ Combined existence of solid-gaseous state of substance.	$(ii)$ Fusion curve
	$(iii)$ Vaporization curve

If liquefied oxygen at $1$ atmospheric pressure is heated from $50\, K$ to $300\, K$ by supplying heat at constant rate. The graph of temperature vs time will be

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