Question

In an industrial process 10kg of water per hour is to be heated from ${20^ \circ }C$ to ${80^ \circ }C$ . To do this, steam at ${150^ \circ }C$ is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at ${90^ \circ }C$ . How many kilograms of steam is required per hour (specific heat of steam $1\,cal\,{g^{ - 1}}{\,^ \circ }{C^{ - 1}}$ , latent heat of vaporization $= 540\,cal\,{g^{ - 1}}$ )?
A. 1g
B. 1kg
C. 10g
D. 10kg

Answer 1

The law of conservation of energy which states that energy can neither be created nor be destroyed. Keeping the law of conservation in mind put the heat energy lost equal to heat energy gained in various conversions and rise and fall of temperature from the situation given. Before evaluating any equation make sure all units are following the same system of units.

Complete step by step solution:
According to law of conservation of energy:
Heat gained in process = Heat lost in process
Expanding the question we can observe the following conversions taking place,
Heat gaining processes:
Heating of 10kg of water from temperature ${20^ \circ }C$ to ${80^ \circ }C$
Heat evolving processes:
Conversion of steam from ${150^ \circ }C$to ${100^ \circ }C$ + Conversion of steam to water + Cooling of water from ${100^ \circ }C$to ${90^ \circ }C$.
Let the mass of steam required per hour to rise temperature of water = m kg
In case of change in temperature of system, energy lost or gained both are given by = $m \times s \times \Delta t$
Where,
m = mass of system whose temperature is changed,
s = specific heat that system and
$\Delta t =$ change in temperature
Putting heat lost and heat gained processes into equation:
Energy gained by system to heat water from ${20^ \circ }C$ to ${80^ \circ }C$= $m \times s \times \Delta t = 10000 \times 1 \times \left( {80 - 20} \right)$
Energy released on cooling of steam from ${150^ \circ }C$to ${100^ \circ }C$ = $m \times s \times \Delta t = 1000m \times 1 \times 50$
Energy released on conversion of steam to water = $1000m \times L$
Where,
L = Latent heat of vaporization
Energy released on cooling of water formed by steam from ${100^ \circ }C$ to ${90^ \circ }C =$ $m \times s \times \Delta t = 1000m \times 1 \times 10$
Putting energy gained equal to energy lost:
Heating of 10kg of water from temperature ${20^ \circ }C$ to ${80^ \circ }C$= Conversion of steam from ${150^ \circ }C$to ${100^ \circ }C$ + Conversion of steam to water + Cooling of water from ${100^ \circ }C$to ${90^ \circ }C$.
Putting statements into equations:
$10000 \times 1 \times \left( {80 - 20} \right) = \left( {1000m \times 1 \times 50} \right) + \left( {1000m \times L} \right) + \left( {1000m \times 10 \times 1} \right)$
Putting value of L from question and evaluating the equation will give:
$1000 \times 1 \times 60 = \left( {1000m \times 1 \times 50} \right) + \left( {1000m \times 540} \right) + \left( {1000m \times 1 \times 10} \right)$
Taking 1000m as a common factor from R.H.S. of equation,
$1000 \times 1 \times 60 = 1000m\left( {\left( {1 \times 50} \right) + \left( {1 \times 540} \right) + \left( {1 \times 10} \right)} \right)$
Evaluating above equation will give:
$60000 = 1000m\left( {600} \right)$
m = 1kg
Steam is required per hour to raise the temperature of 10kg of water from ${20^ \circ }C$ to ${80^ \circ }C$.

Note:
While changing the temperature of any substance and changing the state of the substance, it is very important to differentiate both conditions appropriately. Change in temperature of substance utilizes the heat or evolves the heat which can be observed easily by changing temperature. But when substances undergo change in state it utilizes or evolves the energy but temperature remains constant. This hidden heat that is required to change the state of the substance is called Latent heat of the substance.