What would be the final temperature of a mixture of 50g of water at 20 degree c temperature and 50g of water at 20 degree c temperature?
Answers
Answered by
3
Given mass (M₁) = 50g
Temperature (T₁) = 20° C
Mass (M₂) = 50g
Temperature (T₂) = 20° C
By using the method of mixture ,
Final temperature (T) = (M₁T₁+M₂T₂)/M₁+M₂
= (50×20+50×20)/50+50
= 2000/100
= 20° C .
Temperature (T₁) = 20° C
Mass (M₂) = 50g
Temperature (T₂) = 20° C
By using the method of mixture ,
Final temperature (T) = (M₁T₁+M₂T₂)/M₁+M₂
= (50×20+50×20)/50+50
= 2000/100
= 20° C .
kessrinivas25:
Thank You
Answered by
3
Given mass m1 = 50 grams
Temperature T1 = 20° C
Mass M2 = 50g
Temperature T2 = 20° C
then,
Final Temperature = M1T1 + M2T2 / M1+M2
= (50×20+50×20)/50+50
= 1000+1000/100
= 2000/100
= 20° C .
Therefore, Final Temperature T= 20° C
*Note: In the question, both masses and temperature are equal of same substance. So, when they keep in contact, no change can be seen sincce their temperatures are same. Hence although they keep in contact they attain thermal equilibrium wth the same Temperature i.e., 20° C
Temperature T1 = 20° C
Mass M2 = 50g
Temperature T2 = 20° C
then,
Final Temperature = M1T1 + M2T2 / M1+M2
= (50×20+50×20)/50+50
= 1000+1000/100
= 2000/100
= 20° C .
Therefore, Final Temperature T= 20° C
*Note: In the question, both masses and temperature are equal of same substance. So, when they keep in contact, no change can be seen sincce their temperatures are same. Hence although they keep in contact they attain thermal equilibrium wth the same Temperature i.e., 20° C
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