Answer: Law of conservation of energy
Heat Loss by the metal = Heat Gain by the water (Assume no heat loss to the environment)
M₁C₁ΔT₁= M₂C₂ΔT₂
Where
C₁ and C₂ = the specific heat capacities of the metal and water = C₁?, 4.182J/gK
ΔT₁ and ΔT₂ = Temperature changes of the metal and the water
ΔT₁ = (373 - 300.8)K and ΔT₂= (300.8 - 296.7)K convert to S.I units
M₁ and M₂ are the masses of the metal and water = 59.047g, but
M₂ = density x volume = 1kg/m³ x 0.1m³=0.1kg
Note: Volume of water = 100ml=0.1m³, also density of water = 1kg/m³
We have,
0.059kg x C₁J/KgK (373 - 300.8)K = 0.1kg x 4182J/kgK (300.8 - 296.7)K
4.259C₁=1715J/KgK
C₁=402.5J/KgK
The specific heat of the metal is 402.5J/KgK
Hi there!
For this part, we can disregard the horizontal velocity because horizontal and vertical motion are INDEPENDENT.
We can used the simplified equation to solve for time:
Where:
h = height
g = acceleration due to gravity (assume g = 10 m/s²)
Plug in given values:
t ≈3.162 sec
