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Question 6.PS.4: Transfer of Energy Between Samples by Heating Suppose that y......

Transfer of Energy Between Samples by Heating

Suppose that you have 100. mL H_2O at 20.0 °C and you add to the water 55.0 g iron pellets that had been heated to 425 °C. What is the temperature of both the water and the iron when thermal equilibrium is reached? (Assume that there is no energy transfer to the glass beaker or to the air or to anything else but the water. Assume also that no work is done, that no liquid water vaporizes, and that the density of water is 1.00 g/mL.)

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T_{final} = 42.7 °C

Strategy and Explanation   Thermal equilibrium means that the water and the iron pellets will have the same final temperature, which is what we want to calculate. Consider the iron to be the system and the water to be the surroundings. The energy transferred from the iron is the same energy that is transferred to the water. None of this energy goes anywhere other than to the water. Therefore ΔE_{water} = – ΔE_{iron} and q_{water} = – q_{iron}. The quantity of energy transferred to the water and the quantity transferred from the iron are equal. They are opposite in algebraic sign because energy was transferred from the iron as its temperature dropped, and energy was transferred to the water to raise its temperature.
Specific heat capacities for iron and water are listed in Table 6.1. The mass of water is 100.  mL × 1.00  g/mL = 100. g. T_{initial} for the iron is 425 °C and T_{initial} for the water is 20.0 °C.

q_{water} = – q_{iron}

c_{water} × m_{water} × ΔT_{water} = – c_{iron} × m_{iron} × ΔT_{iron}

(4.184  J  g^{-1} °C^{-1} ) (100.  g) (T_{final} – 20.0 °C) = – (0.451  J g^{-1} °C^{-1}) (55.0  g) (T_{final} – 425 °C)

(418.4  J °C^{-1} ) T_{final} – (8.368 × 10^3  J) = – (24.80  J °C^{-1}1 )T_{final} + (1.054 × 10^4  J )

(443.2  J °C{-1} )T_{final} = 1.891 × 10^4  J

Solving, we find T_{final} = 42.7 °C. The iron has cooled a lot (ΔT_{iron} = – 382 °C) and the water has warmed a little (ΔT_{water} =  22.7 °C).

Reasonable Answer Check  As a check, note that the final temperature must be between the two initial values, which it is. Also, don’t be concerned by the fact that transferring the same quantity of energy resulted in two very different values of ΔT; this difference arises because the specific heat capacities and masses of iron and water are different. There is much less iron and its specific heat capacity is smaller, so its temperature changes much more than the temperature of the water.

Table 6.1   Specific eat Capacities for Some Elements, Compounds, and Common Solids
Substance Specific Heat Capacity ( J g^{-1} °C^{-1})
Elements
Aluminum, Al 0.902
Carbon (graphite), C 0.720
Iron, Fe 0.451
Copper, Cu 0.385
Gold, Au 0.128
Compounds
Ammonia, NH_3(\ell) 4.70
Water (liquid), H_2O(\ell) 4.184
Ethanol, C_2H_5OH(\ell) 2.46
Ethylene glycol (antifreeze), HOCH_2CH_2OH(\ell) 2.42
Water (ice), H_2O(s) 2.06
Carbon tetrachloride, CCl_4(\ell) 0.861
A chlorofluoro- carbon (CFC), CCl_2F_2(\ell) 0.598
Common solids
Wood 1.76
Concrete 0.88
Glass 0.84
Granite 0.79

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