1 kg of water 20-circ-C is- mixed with 800 g of water 80-circ-C- Assuming that no heat is lost to the surroundings- Calculate the final temperature of the mixture-24-44-circ- C46-67-circ- C44-44-circ- C54-44-circ- C

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Toppr Admin · Accepted Answer

Mc-x3B8-1-mc-x3B8-2-mc-x3B8-0since c is constant- assuming water is heated from 0oC-1000-20- - -800-80- -1800-x3B8-020000-64000- 84000-1800-x3B8-0-x3B8-0-46-67oChence the finale temperature in -x3B8-0-46-67o

1 kg of water at $20^{\circ} C$ is, mixed with 800 g of water at $80^{\circ} C$ . Assuming that no heat is lost to the surroundings. Calculate the final temperature of the mixture.
${24.44}^{\circ} C$
${46.67}^{\circ} C$
${44.44}^{\circ} C$
${54.44}^{\circ} C$

$m c θ_{1} + m c θ_{2} = m c θ_{0}$
since c is constant, assuming water is heated from $0^{o} C$
(1000)(20) + (800)(80)= (1800) $θ_{0}$
20000+64000= 84000=1800 $θ_{0}$
$θ_{0} = {46.67}^{o} C$
hence the finale temperature in $θ_{0} = {46.67}^{o}$

1 kg of water at 20∘C is, mixed with 800 g of water at 80∘C. Assuming that no heat is lost to the surroundings. Calculate the final temperature of the mixture.24.44∘C46.67∘C44.44∘C54.44∘C

mcθ1+mcθ2=mcθ0since c is constant, assuming water is heated from 0oC(1000)(20) + (800)(80)= (1800)θ020000+64000= 84000=1800θ0θ0=46.67oChence the finale temperature in θ0=46.67o

1 kg of water at $20^{\circ} C$ is, mixed with 800 g of water at $80^{\circ} C$ . Assuming that no heat is lost to the surroundings. Calculate the final temperature of the mixture.
${24.44}^{\circ} C$
${46.67}^{\circ} C$
${44.44}^{\circ} C$
${54.44}^{\circ} C$

$m c θ_{1} + m c θ_{2} = m c θ_{0}$
since c is constant, assuming water is heated from $0^{o} C$
(1000)(20) + (800)(80)= (1800) $θ_{0}$
20000+64000= 84000=1800 $θ_{0}$
$θ_{0} = {46.67}^{o} C$
hence the finale temperature in $θ_{0} = {46.67}^{o}$