Answer: -112200J
Explanation:
The amount of heat (Q) released from an heated substance depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Since,
Q = ?
Mass of water vapour = 30.0g
C = 187 J/ G°C
Φ = (Final temperature - Initial temperature)
= 100°C - 120°C = -20°C
Then apply the formula, Q = MCΦ
Q = 30.0g x 187 J/ G°C x -20°C
Q = -112200J (The negative sign does indicates that heat was released to the surroundings)
Thus, -112200 joules of heat is released when cooling the superheated vapour.
Answer:
Increase in temperature will not affect a solid - liquid mixture. Example can be taken of salt and water
Condenser Lens - This lens system is located immediately under the stage and focuses the light on the specimen.
Answer:
2300J
Explanation:
1 kilojoule is 1000joules so to get how much is 2.3 multiply it with 1000
