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: 20.775 g S
Explanation: 3.9x10^23 atoms = 0.648 mol
Atomic mass S = 32.08
S in grams = 20.775
<span>E = mCdT
E = energy, m = mass, C = specific heat capacity, dT = change in temperature.
526 = 0.074C x 17
E = 0.074C x 55
Divide the equations
E/526 = (0.074C x 55)/(0.074C x 17) = 55/17
E = (55 x 526)/17 = 1702 J</span>