Question

Water of volume 2L in a closed container is heated with a coil of 1 kW. While water is heated, the container loses energy at a rate of 160 J/s. In how much time will the temperature of water rise from ${27^0}C$ to ${77^0}C$? (Specific heat of water is 4.2 kJ/kg and that of the container is negligible)

A. 8 min 20 s
B. 6 min 2 s
C. 7 min
D. 14 min

Answer 1

It is clearly mentioned in the question that the lid of the container is open. That means the amount of heat heater gives is utilized to raise the temperature and some amount of heat is lost to the surroundings. So we should calculate the amount of heat the heater gives and then equate it to the sum of lost heat and utilized heat.

Formula used:
\eqalign{ & Q = ms\Delta T \cr & {\text{Heat = (power)(time)}} \cr}

Complete step by step answer:
Density of the water is 1 kg/liter that means 1 liter of water weighs one kilogram i.e thousand grams. It’s given 2 liter of water and hence weighs 2000 grams. It is given that the power of the heater is one kilo watt. We can get the heat supplied by this heater in time interval ‘t’. we also have the rate at which the heat is lost and by multiplying this with the time interval ‘t’ we can get the amount of heat lost in ‘t’ seconds, Because ${\text{Heat = (power)(time)}}$.
Now from all the information we have we will find out the time interval required. Let it be ‘t’ seconds
Amount of heat required to raise temperature from ${27^0}C$ to ${77^0}C$ is $Q = mc\Delta T$
\eqalign{ & Q = ms\Delta T \cr & Q = 2 \times 1000 \times 4.2 \times \left( {{{77}^0}C - {{27}^0}C} \right) \cr & Q = 420000J \cr}
The amount of heat lost to the atmosphere is ${\text{Heat = (power)(time) = 160}} \times {\text{t}}$
The amount of heat given by the heater is ${\text{Heat = (power)(time) = }}1kw \times t = (1000t)J$
We equate the amount of heat given by the heater to the heat lost and heat required to raise the temperature.
\eqalign{ & (1000t) = 420000 + {\text{160}} \times {\text{t}} \cr & t = \dfrac{{420000}}{{840}} = 500s = 8\min 20s \cr}

So, the correct answer is “Option A”.

Note:
Lid is open means that this system has become an open system where it can exchange both matter and energy. If it is an isolated system then no heat will be lost to the surroundings and the entire heat given will be utilized to change the temperature of the water. The most important thing is to convert all the given quantities into their SI units.