Question

22. 50 grams of water at 20 °C
is mixed with 50 grams of
water att °C. If the final
temperature is 30 °C. Then find
t°C.
O A. 43 °C
O B. 49 °C
O C. 56 °C
O D. 40 °C​

Answer 1

since it is given that 50 gms water at 20 ⁰ C and 50 gms of water at 20⁰ 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.

0⁰ 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 =

0⁰ 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)

0⁰ 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)

0⁰ 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

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another way using specific heats :

another way using specific heats : let the final temperature be = T ⁰C

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)

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)

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)

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 )

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,

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)

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

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

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 °CSo Your Answer is Option (D) Which is 40 °C

Answer 2

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.

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final temperature of the mixture =

= [ m1 * T1 + m2 * T2 ] / (m1 + m2)

= [ 50 gms * 20⁰ C + 50 gms * 40⁰C ] / (50+50)

= 3,000 / 100 = 30⁰C

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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 = 30⁰C