# Termodinamica

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• Publicado : 15 de febrero de 2011

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Temperature 1-26C The zeroth law of thermodynamics states that two bodies are in thermal equilibrium if both have the
same temperature reading, even if they are not in contact.
1-27C They arecelsius(°C) and kelvin (K) in the SI, and fahrenheit (°F) and rankine (R) in the English system.
1-28C Probably, but not necessarily. The operation of these two thermometers is based on thethermal
expansion of a fluid. If the thermal expansion coefficients of both fluids vary linearly with temperature, then both fluids will expand at the same rate with temperature, and both thermometerswill always give identical readings. Otherwise, the two readings may deviate.

1-29 A temperature is given in °C. It is to be expressed in K. Analysis The Kelvin scale is related to Celsius scaleby
T(K] = T(°C) + 273 Thus, T(K] = 37°C + 273 = 310 K

1-30E A temperature is given in °C. It is to be expressed in °F, K, and R. Analysis Using the conversion relations between the varioustemperature scales,
T(K] = T(°C) + 273 = 18°C + 273 = 291 K T(°F] = 1.8T(°C) + 32 = (1.8)(18) + 32 = 64.4°F T(R] = T(°F) + 460 = 64.4 + 460 = 524.4 R

1-31 A temperature change is given in°C. It is to be expressed in K. Analysis This problem deals with temperature changes, which are identical in Kelvin and Celsius scales.
Thus, ∆T(K] = ∆T(°C) = 15 K
1-32E A temperature change isgiven in °F. It is to be expressed in °C, K, and R. Analysis This problem deals with temperature changes, which are identical in Rankine and Fahrenheit
scales. Thus, ∆T(R) = ∆T(°F) = 45 R
Thetemperature changes in Celsius and Kelvin scales are also identical, and are related to the changes in Fahrenheit and Rankine scales by
∆T(K) = ∆T(R)/1.8 = 45/1.8 = 25 K and ∆T(°C) = ∆T(K) = 25°C1-33 Two systems having different temperatures and energy contents are brought in contact. The direction
of heat transfer is to be determined. Analysis Heat transfer occurs from warmer to cooler...