Macmillan Learning You heat a 5.2 gram lead ball (specific heat capacity=0.128 Joules/g-deg) to 183. "C. You then drop the ball into 34.5 milliliters of water (density 1 g/ml; specific heat capacity = 4.184 J/g-deg) at 22.4 °C. What is the final temperature of the water when the lead and water reach thermal equilibrium? temperature: °C

Chemistry by OpenStax (2015-05-04)
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Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
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Chapter5: Thermochemistry
Section: Chapter Questions
Problem 31E: When a 0.740-g sample of trinitrotoluene (TNT), C7H5N2O6, is burned in a bomb calorimeter, the...
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You heat a 5.2 gram lead ball (specific heat capacity = 0.128 Joules/g-deg) to 183. °C. You then drop the ball into
34.5 milliliters of water (density 1 g/ml; specific heat capacity = 4.184 J/g-deg) at 22.4 °C. What is the final temperature of the
water when the lead and water reach thermal equilibrium?
temperature:
▬▬▬
O Search
Resources Q
Insert
°C
Delete
Transcribed Image Text:You heat a 5.2 gram lead ball (specific heat capacity = 0.128 Joules/g-deg) to 183. °C. You then drop the ball into 34.5 milliliters of water (density 1 g/ml; specific heat capacity = 4.184 J/g-deg) at 22.4 °C. What is the final temperature of the water when the lead and water reach thermal equilibrium? temperature: ▬▬▬ O Search Resources Q Insert °C Delete
If you have 0.366 m³ of water at 25.0 °C in an insulated container and add 0.132 m³ of water at 95.0 °C, what is the final
temperature Tr of the mixture? Use 1000 kg/m³ as the density of water at any temperature.
QT₁=
T
O Search
08
*-
F11
☀+
F12
PrtSc
Resources
Insert
Delete
°C
Transcribed Image Text:If you have 0.366 m³ of water at 25.0 °C in an insulated container and add 0.132 m³ of water at 95.0 °C, what is the final temperature Tr of the mixture? Use 1000 kg/m³ as the density of water at any temperature. QT₁= T O Search 08 *- F11 ☀+ F12 PrtSc Resources Insert Delete °C
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