A new temperature scale uses X as a unit of t | vedclass

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Question

(A) Given the relation between the new scale $t_x$ and absolute temperature $T$ is $t_x = 3T + 100$.The change in temperature $\Delta t_x$ is related

Vedclass · Accepted Answer

$4200 \, J \, kg^{-1} K^{-1}$

Vedclass · Answer

(A) Given the relation between the new scale $t_x$ and absolute temperature $T$ is $t_x = 3T + 100$.The change in temperature $\Delta t_x$ is related to the change in absolute temperature $\Delta T$ by:$\Delta t_x = t_{x2} - t_{x1} = (3T_2 + 100) - (3T_1 + 100) = 3(T_2 - T_1) = 3 \Delta T$.Specific heat $c$ is defined as the amount of heat $Q$ required to raise the temperature of unit mass $m$ by unit temperature change $\Delta 	heta$:$c = \frac{Q}{m \Delta 	heta}$.Given $c_x = 1400 \, J \, kg^{-1} X^{-1}$,we have $c_x = \frac{Q}{m \Delta t_x} = 1400$.We want to find $c_{SI} = \frac{Q}{m \Delta T}$.Substituting $\Delta t_x = 3 \Delta T$ into the expression for $c_x$:$1400 = \frac{Q}{m (3 \Delta T)} = \frac{1}{3} \left( \frac{Q}{m \Delta T} ight) = \frac{1}{3} c_{SI}$.Therefore,$c_{SI} = 3 	imes 1400 = 4200 \, J \, kg^{-1} K^{-1}$.

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