Change in temperature = final temperature - Initial temperature
Δt = t₂ - t₁
Δt = 17 - (-6)
Δt = 17 + 6 = 23 f
In short, Your Answer would be Option D
Hope this helps!
Answer:
Q=mc(T2-T1)
Explanation:
Ti is the temperature
m is mass
c is specific heat capacity for steam
Q is heat, [Q]=J
Scientific evidence supports a new idea
Answer:
0.267 × 10⁻³
Explanation:
Given:
maximum permissible workday dose = 18 mrem
Weight of the laboratory technician = 54 kg
gamma rays absorbed = 2.6 mJ of 0.30 MeV
relative biological effectiveness (RBE) for gamma rays = 1.00
Now,
Radiation Absorbed Dose 
also,
Roentgen Equivalent Man = rad × Q
here Q is the quality factor = 1 for gamma rays
thus,
Roentgen Equivalent Man = 4.8 × 10⁻³ × 1 = 4.8 × 10⁻³ rem
Therefore,
the ratio of the equivalent dosage received by the technician to the maximum permissible equivalent dosage will be
= 
or
= 0.267 × 10⁻³