10 kJ താപം ആഗിരണം ചെയ്യുമ്പോൾ 80 ° C താപനില വർദ്ധിക്കുന്ന 500 ഗ്രാം മാസുള്ള ലോഹ ബ്ലോക്കിന്റെ വിശിഷ്ട താപധാരിത (Jg-1K-1 ൽ) കണ്ടെത്തുക ?

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RRB ALP Fitter 23 Jan 2019 Official Paper (Shift 1)
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  1. 0.4
  2. 0.16
  3. 1.56
  4. 0.25

Answer (Detailed Solution Below)

Option 4 : 0.25
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ആശയം:

വിശിഷ്ട താപധാരിത:

  • ഒരു വസ്തുവിന്റെ താപനില ഒരു ഡിഗ്രി മാറ്റം വരുത്താൻ ആവശ്യമായ താപത്തിന്റെ അളവ് എന്ന് നിർവചിക്കാം.
  • ഗണിതശാസ്ത്രപരമായി, ആവശ്യമായ താപം, Q = mcΔT,
  • ഇവിടെ m = മാസ്, c =വിശിഷ്ട താപധാരിത, ΔT = താപനിലയിലുള്ള വ്യത്യാസം
  • ഇത് Jg-1K-1-ൽ അളക്കാം.

ഗണിതക്രിയ:

തന്നിരിക്കുന്നത്: മാസ്, m = 500 g, താപനിലയുള്ള വ്യത്യാസം, ΔT = 80ºC,താപം, Q = 10 kJ

വിശിഷ്ട താപധാരിത\(c=\frac{Q}{mΔ T}\)

\(c=\frac{10\times 10^3}{500\times 80}=0.25 ~Jg^{-1}K^{-1}\)

അതിനാൽ,വിശിഷ്ട താപധാരിത 0.25 Jg-1K-1 ആണ്.

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