+ 2.51 x 102 J/°C)-9.01 x 104 T- 37.99c LEARN MORE REMARKS The answer turned out

Question

+ 2.51 x 102 J/°C)-9.01 x 104 T- 37.99c LEARN MORE REMARKS The answer turned out to be very close to the aluminum's initial temperature, so it would have been impossible to guess in advance whether the aluminum would lose or gain energy. Notice the way the table was organized, mirroring the order of factors in the different terms. This kind of organization helps prevent substitution errors, which are common in these problems. QUESTION Suppose thermal energy 0 leaked from the system. How should the right side of Equation (1) be adjusted? No change is needed. PRACTICE IT Use the worked example above to help you solve this problem. Suppose 0.385 kg of water initially at 43.5°C is poured into a 0.300 kg glass beaker having a temperature of 25.0°C. A 0.500 kg block of aluminum at 37.0°C is placed in the water, and the system insulated. Calculate the final equilibrium temperature of the system oc EXERCISE HINTS: GETTING STARTED I IM STUCK! A 19.5 kg gold bar at 28.0°C is placed in a large, insulated 0.800 kg glass container at 15.0°C with 2.00 kg of water at 25.0°c. Calculate the final equilibrium temperature oC Need Help? Talk to a Tutor

Explanation / Answer

First two problems are related to the another one which is not given, I'm solving last one...

Q = m*c*dT

m = mass, c = specific heat

Heat lose = heat gain

gold specific heat = 0.129 J/g K

water specific heat = 4.186 J/g K

glass specific heat = 0.15

19.5 x 10^3 g * 0.129 * (28 - T) = 0.800 x 10^3 * 0.15 * (T -15) + 2 x 10^3 * 4.186 * (T - 25)

70434 - 2515.5T = 120T - 1800 + 8372T - 209300

280276.25 = 11007.5T

T = 25.58 oC

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