मैक्सिमम इज यूज टू री सा इस द विंडो बैक टू इट
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Answer:
Explanation:
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CBSE/Dir/Sk. Ed.//2020 Date : April 4th, 2020
Circular No. : Acad-24/2020
Dear Principals,
The entire world is facing a completely unprecedented crisis due to the spread of
the Novel Corona Virus. Social distancing, one of the key components in the fight
with this virus, is creating a challenging situation for our social institutions. Schools
as prominent social institutions, are not going to remain unaffected by it. You, as
Principals, have a huge role to play in the present setting. Yes, you have your
anxieties, and fairly so. Yet there is a tremendous window of opportunity to rethink
many things. You can in fact, reinvent the education processes of your school,
devise creative methods to engage with your stakeholders, and perhaps even set
examples to follow for the entire education system. We believe that this is the time
to not just deal with the current emergency, but also build foundations of resilience
to face such difficult events in the future. Here are a few thoughts and suggestions
for your school, which you might like to take up voluntarily, of course by keeping in
mind the local limitations.
It takes a village to raise a child:
It’s the beginning of a new academic session, but with academic days getting lost,
you may be wondering how you will catch up when the country overcomes the
present situation. Don’t let this pressure and disquiet consume you. Rather, use this
time to leverage certain untapped resources. What about getting in touch with those
enthusiasts in civil society who are constantly working on some or the other
intervention for ensuring quality in education? Can the home be engaged by the
school for some lateral solutions? Can outfits that have built high credibility around
integrating education with art, music, cinema, quizzes, gardening, informal setups
and so on, be summoned for preparing handy toolboxes? Use this time to tap on all
these resources and communities to develop creative and innovative teaching and
learning material, such as - activity-based worksheets; projects that can be done at
home; activities that can generate a flood of curiosity; art, sport and music-
integrated learning modules; online classes to build capacities of your teachers to
open up their minds to creativity, etc. The sky is the limit here.
Covering the course:
The lockdown is building new arteries within homes; families are rediscovering their
bonds and connections; they are making renewed efforts to do things that they
enjoy together as a family. Therefore, how about considering this challenge as an
opportunity to shift focus from ‘schooling only at schools’ to ‘School-Home
collaboration for learning’. Let us relook at the various symbiotic relationships –
Answer:
since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 40⁰ C are mixed. Since the masses of the liquid at different temperatures are same, the answer is very easy and simple : average of 20⁰C and 40⁰C. that is: 30⁰C.
=====================
final temperature of the mixture =
= [ m1 * T1 + m2 * T2 ] / (m1 + m2)
= [ 50 gms * 20⁰ C + 50 gms * 40⁰C ] / (50+50)
= 3,000 / 100 = 30⁰C
====================
another way using specific heats :
let the final temperature be = T ⁰C
Amount of heat given out by the hot water = m * s * (40⁰C - T)
= 50 gms * s* (40 -T)
Amount of heat taken in by the cold water = m * s * (T - 20⁰C)
= 50 gms * s * (T - 20 )
As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,
50 * s * (40 -T) = 50 gm * s * (T-20)
40 - T = T - 20
2 T = 60 => T = 40 °C
So Your Answer is Option (D) Which is 40 °C
since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 40⁰ C are mixed. Since the masses of the liquid at different temperatures are same, the answer is very easy and simple : average of 20⁰C and 40⁰C. that is: 30⁰C.
=====================
final temperature of the mixture =
= [ m1 * T1 + m2 * T2 ] / (m1 + m2)
= [ 50 gms * 20⁰ C + 50 gms * 40⁰C ] / (50+50)
= 3,000 / 100 = 30⁰C
====================
another way using specific heats :
let the final temperature be = T ⁰C
Amount of heat given out by the hot water = m * s * (40⁰C - T)
= 50 gms * s* (40 -T)
Amount of heat taken in by the cold water = m * s * (T - 20⁰C)
= 50 gms * s * (T - 20 )
As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,
50 * s * (40 -T) = 50 gm * s * (T-20)
40 - T = T - 20
2 T = 60 => T = 40 °C
So Your Answer is Option (D) Which is 40 °C
since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 40⁰ C are mixed. Since the masses of the liquid at different temperatures are same, the answer is very easy and simple : average of 20⁰C and 40⁰C. that is: 30⁰C.
=====================
final temperature of the mixture =
= [ m1 * T1 + m2 * T2 ] / (m1 + m2)
= [ 50 gms * 20⁰ C + 50 gms * 40⁰C ] / (50+50)
= 3,000 / 100 = 30⁰C
====================
another way using specific heats :
let the final temperature be = T ⁰C
Amount of heat given out by the hot water = m * s * (40⁰C - T)
= 50 gms * s* (40 -T)
Amount of heat taken in by the cold water = m * s * (T - 20⁰C)
= 50 gms * s * (T - 20 )
As the amounts are equal, because the heat is transferred from hotwater to the cold water without any loss of heat to any surroundings,
50 * s * (40 -T) = 50 gm * s * (T-20)
40 - T = T - 20
2 T = 60 => T = 40 °C
So Your Answer is Option (D) Which is 40 °C