Question 1000: Thermodynamics & KTG

Question

An electrically heated coil is immersed in a calorimeter containing $360 \mathrm{~g}$ of water at $10^{\circ} \mathrm{C}$. The coil consumes energy at the rate of $90 \mathrm{~W}$. The water equivalent of calorimeter and coil is $40 \mathrm{~g}$. The temperature of water after $10 \mathrm{~min}$ is

Accepted Answer

Energy supplied by the heater to the system in $10 \mathrm{~min}$
$\begin{aligned}
Q_1 & =P 	imes t=90 \mathrm{~J} / \mathrm{s} 	imes 10 	imes 60 \mathrm{~s} \
& =54000 \mathrm{~J}=\frac{54000}{42} \mathrm{cal} \
& =12857 \mathrm{cal}
\end{aligned}$
Now, if $	heta$ is the final temperature of the system, energy absorbed by it to change its temperature from $10^{\circ} \mathrm{C}$ to $	heta^{\circ} \mathrm{C}$,
$\begin{aligned}
& Q_2=(m s \Delta T)_{	ext {water }}+(m s \Delta T)_{	ext {coil + calorimeter }} \
&=360 	imes 1 	imes(	heta-10)+40(	heta-10) \
&=400(	heta-10)
\end{aligned}$
According to question,
$Q_1=Q_2$
$\begin{aligned}
& 	herefore & 12857 & =400^{\circ}(	heta-10) \
& 	ext { or } & \quad 	heta & =42.14^{\circ} \mathrm{C}
\end{aligned}$

Answer

$4.214^{\circ} \mathrm{C}$

Answer

$42.14^{\circ} \mathrm{C}$

Answer

$30^{\circ} \mathrm{C}$

Answer

$45.18^{\circ} \mathrm{C}$

Thermodynamics & KTG

An electrically heated coil is immersed in a calorimeter containing 360 \mathrm{~g} of water at 10^{\circ} \mathrm{C}. The coil consumes energy at the rate of 90 \mathrm{~W}. The water equivalent of calorimeter and coil is 40 \mathrm{~g}. The temperature of water after 10 \mathrm{~min} is

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