Given that:
Mass (m) = 20 g = 0.02 Kg,
temperature (T₁) = -10°C = -10+273 = 263 K
temperature (T₂) = 10°C = 10+273 = 283 K
Specific heat of water (Cp) = 4.187 KJ/Kg k
We know that Heat transfer (Q) = m. Cp.( T₂ - T₁)
= 0.02 × 4.187 × (283-263)
<em> Q = 1.67 KJ</em>
<em>Heat transferred is 1.67 KJ</em>
Answer: D) It should be handled in a fume hood, away from open flames.
Answer:
W = 250(50) = 12500 J
Explanation:
it would take him 1 minute to run 304.5 meters and 1 second to run 5.075 meters