A 20.0 g sample of aluminum (specific heat = 0.902) g-1 oC-1) with an initial temperature of 2.5°C is heated with 427 J of energy. What is the final temperature of the sample?
O 72.3°C
O 23.7°C
0 26.2°C
0 74.8°C
O 24.9°C

T(final) = 26.2°C (3 sig. figs.)

Explanation:

Q = m·c·ΔT

m = mass of sample of interest = 20.0g

c = specific heat of sample of interest = 0.902j·g⁻¹·C⁻¹

ΔT = Temperature change = T(final) - T(initial) = T(f) - 2.5°C

Q = 427 joules

Q = m·c·ΔT => = ΔT = Q/m·c => T(f) = Q/m·c + T(i) = (427j/20.0g·0.902j·g⁻¹·C⁻¹) + 2.5°C = 26.16962306°C (calc. ans.) ≅ 26.2°C (3 sig. figs.)

it can be observed without changing the identity of the object.

it depends on what substance you're talking about as well as many other factors. in quantum mechanics, the observation of particles affects the observers ability to measure certain properties. for example: you can know the velocity of a particle, but you can't know it's position; or you can know it's position, but not it's velocity. it's like trying to find out where a grain of sand is by throwing other grains of sand at it. once you find out where it is, it's been moved. it's also like trying to find an answer to the question of whether or not a tree falling in a forest makes a sound if nobody is there to hear it. paradoxically, the only way to find out is to be there and hear it.