Homework Help: physics

Posted by Akbar on Thursday, April 26, 2012 at 4:41pm.

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?

I need help on this please could someone please help out.Thank you!

Physics - bobpursley, Tuesday, April 24, 2012 at 5:53pm

the sum of heats gained is zero.

massglass*specheatglass*(21.6-38.7)+135*specheatwater*(Ti-38.6)=0

solve for Ti

Physics - Akbar, Tuesday, April 24, 2012 at 6:58pm

I got 15.43 degree Celsius but its wrong. I don't know why please help. Thank you!

physics - bobpursley, Wednesday, April 25, 2012 at 12:14pm

well, it is wrong, you know the final answer is between 21.6 and 38.7C

Let me see your calculations, including what you used as specific heats of glass and water.

physics - Akbar, Wednesday, April 25, 2012 at 12:36pm

This is how I did it:

(.034)(840)(21.6-38.7)+135(4186)(Ti-38.6)

physics - Elena, Thursday, April 26, 2012 at 6:48pm
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

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