Answer
For Tin, 202 K or -71 °C
For Molybdenum, 1156 K or 883°C
For Iron, 724 K or 451 °C
For Gold, 535 K or 262 °C
For Zinc, 277 K or 4 °C
For Chromium, 859 K or 586 °C
Work Step by Step
Required:
Approximate temperature at which creep deformation becomes an important consideration for each of the following metals (melting temperatures can be found from the cover of the book).
Solution:
Creep becomes important at about $0.4T_{m},$ where $T_{m}$ is the absolute melting temperature of metal. Solving for the approximate temperature:
For Tin, $0.4T_{m} = (0.4)(232 + 273) $= 202 K or -71 °C
For Molybdenum, $0.4T_{m} = (0.4)(2617 + 273) $= 1156 K or 883°C
For Iron, $0.4T_{m} = (0.4)(1538 + 273) $= 724 K or 451 °C
For Gold, $0.4T_{m} = (0.4)(1064 + 273) $= 535 K or 262 °C
For Zinc, $0.4T_{m} = (0.4)(420 + 273) $= 277 K or 4 °C
For Chromium, $0.4T_{m} = (0.4)(1875 + 273) $= 859 K or 586 °C
