Answer: The scientist gives up and starts an investigation on a new topic.
Explanation:
The data is altered so that it supports the original hypothesized. The data is then altered so that it supports the original hypothesis.
Answer:Poopy-di scoop
Scoop-diddy-whoop
Whoop-di-scoop-di-poop
Poop-di-scoopty
Scoopty-whoop
Whoopity-scoop, whoop-poop
Poop-diddy, whoop-scoop
Poop, poop
Scoop-diddy-whoop
Whoop-diddy-scoop
Whoop-diddy-scoop, poop
Explanation:
The answer they are looking for is the last one. However the last two are technically correct but the third one would result in negative work.
Thermal conductions
K= QL/ART
Aluminium T₁ = 10 + 273.15
T₂ = 283.15k
205 = 2.0 × 0.30/4× 10⁻⁴ × (T₂ - 283.15)
Copper
385 = Q × 0.70/4×10⁻⁴ ×(433.15 - T₂)
Where T₃ = 160 + 273.15
T₃ = 433.15K
From 2 to 3
205/385 = 0.30/0.70 × 433.15 - T₂/T₂ - 283.15
= 0.53T₂ -150.06 = 181.92 - 0.42 T₂
→ 0.95T₂ = 331.98 ⇒ T₂ = ₂349.45k
T₂ = 76.3°c
=77°c.
Answer:
Explanation:
When 2 gms of steam condenses to water at 100 degree latent heat of vaporization is releases which is calculated as follows
Heat released = mass x latent heat of vaporization
= 2 x 2260 = 4520 J
When 2 gms of water at 100 degree is cooled to ice water at zero degree heat is releases which is calculated as follows
Heat released = mass x specific heat x( 100-0)
= 2 x 4.2 x 100 = 840 J
When 2 gms of water at zero degree condenses to ice at zero degree latent heat of fusion is releases which is calculated as follows
Heat released = mass x latent heat of fusion
= 2 x 334 = 668 J
When 2 grams of steam at 100 degrees Celsius turns to ice at 0 degrees Celsius heat released will be sum of all the heat released as mentioned above ie
4520 + 840 +668 = 6028 J