An aluminium container of mass 100 , g contai | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Total heat supplied in $4$ minutes $(240 \, s)$ is $Q = 100 \, cal/s 	imes 240 \, s = 24000 \, cal$.Step $1$: Heat required to raise the temperat

Vedclass · Accepted Answer

$25.5$

Vedclass · Answer

(B) Total heat supplied in $4$ minutes $(240 \, s)$ is $Q = 100 \, cal/s 	imes 240 \, s = 24000 \, cal$.Step $1$: Heat required to raise the temperature of the system from $-20^{\circ} C$ to $0^{\circ} C$:$Q_1 = m_{Al} c_{Al} \Delta T + m_{ice} c_{ice} \Delta T = (100 	imes 0.2 	imes 20) + (200 	imes 0.5 	imes 20) = 400 + 2000 = 2400 \, cal$.Time taken for this step $= 2400 / 100 = 24 \, s$.Step $2$: Heat required to melt the ice at $0^{\circ} C$:$Q_2 = m_{ice} L = 200 	imes 80 = 16000 \, cal$.Time taken for this step $= 16000 / 100 = 160 \, s$.Total time elapsed $= 24 + 160 = 184 \, s$. Remaining time $= 240 - 184 = 56 \, s$.Remaining heat available $= 24000 - (2400 + 16000) = 5600 \, cal$.Step $3$: Heat used to raise the temperature of the system (water + container) from $0^{\circ} C$ to $	heta$:$Q_3 = (m_{Al} c_{Al} + m_{water} c_{water}) \Delta 	heta = (100 	imes 0.2 + 200 	imes 1) 	heta = (20 + 200) 	heta = 220 	heta$.$220 	heta = 5600 \implies 	heta = 5600 / 220 \approx 25.45^{\circ} C \approx 25.5^{\circ} C$.

Similar Questions

Water falls from a height of $500 \ m$. What is the rise in temperature of water at the bottom if the whole energy remains in the water? (Assume $g = 9.8 \ m/s^2$ and specific heat capacity of water $c = 4200 \ J/kg \cdot ^\circ C$)

If $1\; g$ of steam at $100^{\circ}C$ is mixed with $1\; g$ of ice at $0^{\circ}C$,then the resultant temperature of the mixture is ........ $^{\circ}C$.

The freezing point of a liquid decreases when pressure is increased,if the liquid:

Vedclass Test Series

Exam Paper Generator

Online Exam Module