11. 0.00 kJ 1.02 × 10 5 kJ 4.10 × 10 6 kJ 1.50 × 10 4 kJ 12. A temperature often
ID: 1999216 • Letter: 1
Question
11.
0.00 kJ
1.02 × 105kJ
4.10 × 106kJ
1.50 × 104 kJ
12.
A temperature often called absolute zero is approximately -273°C. What is this temperature in kelvins?
-273 K
0 K
273 K
18 K
13.
The addition of heat Q causes a metal object to increase in temperature from 4°C to 6°C . What is the amount of heat necessary to increase the object's temperature from 6°C to 12°C?
4Q
2Q
Q
3Q
0.00 kJ
1.02 × 105kJ
4.10 × 106kJ
1.50 × 104 kJ
12.
A temperature often called absolute zero is approximately -273°C. What is this temperature in kelvins?
-273 K
0 K
273 K
18 K
13.
The addition of heat Q causes a metal object to increase in temperature from 4°C to 6°C . What is the amount of heat necessary to increase the object's temperature from 6°C to 12°C?
4Q
2Q
Q
3Q
f 334 kJ/kgExplanation / Answer
11. In this case the temperature remains same. That means we have to melt 45kg of ice of 0 degree C into 0 degree C water. The heat required Q=M *Lf = 45* 334 KJ = 1.5 *10^4 KJ
12. To convert centigrade to Kelvin we use the following formula,
T(K) = T(C) +273
Therefore for -273 degree C we get T(K) = -273 +273= 0K
13. We consider the mass of the metal as m and s as specific heat. At 4 degree C the initial heat was Q0 . To reach 6 degree C the metal needs additional heat Q
Therefore at 6 degree C total heat in the metal is Q+ Q0. Using basic formula Q+ Q0 = m*s * (T2-T1)
or Q+ Q0 = m*s * (6-4) = 2*m*s ....................... (1)
Now to increase the temperature from 6 to 12 degree the amount of heat required H= m*s *(T3-T2)
or H= m*s (12-6) =m*s*6 ................................. (3)
from equation (1) we get, m*s = [ Q+ Q0]/2 ..................... (2)
using equation (2) in equation (3) we get,
H= 6* [ Q+ Q0]/2 = 3[Q+ Q0]
using m, s and Q0, Q for the metal we get the heat required.
Answer will be 3Q
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