800 calories of heat is required to raise the temperature of 0.080 18
liquid from 10 °C to 100 °C. Find its specific capacity in cal kg .
1) 111.1
2) 2.222
3) 11.11
4) 2222
Answers
Answered by
0
Answer:
2 ) 2.222
Step-by-step explanation:
option 2
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Answered by
2
Specific heat capacity of the liquid is 1) 111.1 cal
Mass of the liquid = M = 0.080 Kg (Given)
Heat supplied = Q = 800 calories. (Given)
Initial temperature = 10°C (Given)
Final Temperature = 100°C.(Given)
Therefore,
Temperature difference = 100° - 10° = 90°
Now,
Q = mcT, where c is the specific heat
c = Q / mΔT
c = 800/ 0.080 × 90
c = 800 / 7.2 cal
c = 111.1
Thus, the specific heat capacity is 111.1 kg
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