Answer:
The specific heat capacity of iridium = 0.130 J/g°C
Explanation:
Assuming no heat losses to the environment and to the calorimeter,
Heat lost by the iridium sample = Heat gained by water
Heat lost by the iridium sample = mC ΔT
m = mass of iridium = 23.9 g
C = specific heat capacity of the iridium = ?
ΔT = change in temperature of the iridium = 89.7 - 22.6 = 67.1°C
Heat lost by the iridium sample = (23.9)(C)(67.1) = (1603.69 C) J
Heat gained by water = mC ΔT
m = mass of water = 20.0 g
C = 4.18 J/g°C
ΔT = 22.6 - 20.1 = 2.5°C
Heat gained by water = 20 × 4.18 × 2.5 = 209 J
Heat lost by the iridium sample = Heat gained by water
1603.69C = 209
C = (209/1603.69) = 0.130 J/g°C
I hope it helps! good luck in chem!
hmax = 5740.48 m. The maximum height that a cannonball fired at 420 m/s at a 53.0° angles is 5740.48m.
This is an example of parabolic launch. A cannonball is fired on flat ground at 420 m/s at a 53.0° angle and we have to calculate the maximum height that it reach.
V₀ = 420m/s and θ₀ = 53.0°
So, when the cannonball is fired it has horizontal and vertical components:
V₀ₓ = V₀ cos θ₀ = (420m/s)(cos 53°) = 252.76 m/s
V₀y = V₀ cos θ₀ = (420m/s)(cos 53°) = 335.43m/s
When the cannoball reach the maximum height the vertical velocity component is zero, that happens in a tₐ time:
Vy = V₀y - g tₐ = 0
tₐ = V₀y/g
tₐ = (335.43m/s)/(9.8m/s²) = 34.23s
Then, the maximum height is reached in the instant tₐ = 34.23s:
h = V₀y tₐ - 1/2g tₐ²
hmax = (335.43m/s)(34.23s)-1/2(9.8m/s²)(34.23s)²
hmax = 11481.77m - 5741.29m
hmax = 5740.48m
<span>.Ask a Question
.Do Background Research.
.Construct a Hypothesis.
.Test Your Hypothesis by Doing an Experiment
.Analyze Your Data and Draw a Conclusion.
<span>.Communicate Your Results.</span></span>
