Asked by Akbar
A 34 g glass thermometer reads 21.6°C before it is placed in 135 mL of water. When the water and thermometer come to equilibrium, the thermometer reads 38.7°C. What was the original temperature of the water?
-.135(4186)(Tf-38.7)=.034(840)(Tf-21.6)
=-565.11Tf+21869.752=28.56Tf-616.896
solve for Tf
Tf= 37.88 degree celcius but its wrong please help!
thank you
-.135(4186)(Tf-38.7)=.034(840)(Tf-21.6)
=-565.11Tf+21869.752=28.56Tf-616.896
solve for Tf
Tf= 37.88 degree celcius but its wrong please help!
thank you
Answers
Answered by
Elena
m1•c1•(38.7-21.6) = m2•c2•(t-38.7)
m2 =ρ•V =1000• 135•10^-3•10^-3 =0.135 kg,
c1 =840 J/kg , c2 = 4180 J/kg.
t -38.7 = 0.034•840•17.1/0.135•4180 = 0.865 oC.
t = 38.7+ 0.865 =39.57oC
m2 =ρ•V =1000• 135•10^-3•10^-3 =0.135 kg,
c1 =840 J/kg , c2 = 4180 J/kg.
t -38.7 = 0.034•840•17.1/0.135•4180 = 0.865 oC.
t = 38.7+ 0.865 =39.57oC
Answered by
John
A 30 g glass thermometer reads 21.6°C before it is placed in 135 mL of water. When the water and thermometer come to equilibrium, the thermometer reads 38.1°C. What was the original temperature of the water?
Please provide step by step equation and answer
Please provide step by step equation and answer
Answered by
M
If no specific heat capacity of gladd
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