ACTIVITY - 7
The entire poem is an excellent example of PERSONIFICATION
(Inanimate objects are spoken of as having life or given human
qualities)
Eg. I tumble down
Water is given the human quality of tumbling down
Another example
I blind the sky with lightning
Identify the lines in the poem in which personification is used.
Answers
Explanation:
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
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