Question

2 litre water at ${27^0}C$ is heated by a $1\,\,kW$ heater in an open container. On an average heat is lost to surroundings at the rate $160J/s.$ The time required for the temperature to reach ${77^0}C$ is
A. $8\,\,\min \,\,18\,\,\sec $
B. $10\,\,\min \,$
C. $7\,\,\min \,$
D. $14\,\,\min \,$

Answer 1

Hint:1. One litre of water has a mass of almost exactly one kilogram $\left( {1kg = 1000gram} \right)$ when measured at its maximal density, which occurs at about ${4^0}C$.
2. If m is the mass of material (here water) in gram for raising the temperature $\Delta t$ the magnitude of heat supplied is given by $\Delta Q = ms\Delta t$
Where s is the specific heat capacity of the material.
3. $S = 1$ Calorie$/g - {}^0C$for water and $S = \dfrac{1}{2}$ Calorie$/g - {}^0C$for ice.
4. For converting $1$Calorie into Joule, we multiply it by $4.18.$
Because, $1\,\,Calorie = 4.18\,\,Joule.$

Complete Step by Step Answer:
The heat required to raise the temperature of $2000gm$ of water from ${27^0}C$to ${77^0}C$
$\Delta Q = ms\Delta t$
Given, $m = 2000gm$
We know the specific heat capacity for water is $s = 1$
Calorie/\[gm - {}^0C\]
Hence, $\Delta Q = 2000 \times 1 \times \left( {77 - 27} \right)$
$ = 2000 \times 50$
$\Delta Q = 100000$
Or, $\Delta Q = {10^5}Calorie$
$\Delta Q = {10^5} \times \left( {1\,\,Calorie} \right)$
$ = {10^5} \times \left( {4.18\,\,Joule} \right)$
$\left[ {\because 1\,\,Calorie = 4.18\,\,Joule} \right]$
$\Delta Q = 4.18 \times {10^5}Joule$
Power given by the heater is
$ = 1KW = 1000W$
$ = 1000\,\,Joule/second$
Now, Total power given by the heater is $1000J/\operatorname{s}$, in which $160J/s$ power gets lost. So the remaining power would be
$ = 1000 - 160$
$P = 840J/S$
So time required for the temperature to reach $t = \dfrac{{heat\,\,given}}{{remaining\,\,power}} = \dfrac{{\Delta Q}}{P}$
$t = \dfrac{{4.18 \times {{10}^5}}}{{840}}\sec ond$
$t \simeq 498$second $ \simeq 8\,\,\min ute\,\,18\,\,\sec ond$
Hence, option (A) is correct.

Note:If power generate by heater is `P’ watt, how some power say $`{P_1}'$ watt is lost into surrounding then the remaining power or power available will be ${P_R} = P - {P_1}.$ Now, this power has basically been used to raise the temperature of the material (here material is water).