Gaussian surfaces A and B enclose the same positive charge+Q. The area of Gaussian surface A is three times larger than that of
1 answer:
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Answer:
= 17º C
Explanation:
This is a calorimetry problem, where heat is yielded by liquid water, this heat is used first to melt all ice, let's look for the necessary heat (Q1)
Let's reduce the magnitudes to the SI system
Ice m = 80.0 g (1 kg / 1000 g) = 0.080 kg
L = 3.33 105 J / kg
Water M = 860 g = 0.860 kg
= 4186 J / kg ºC
Q₁ = m L
Q₁ = 0.080 3.33 10⁵
Q₁ = 2,664 10⁴ J
Now let's see what this liquid water temperature is when this heat is released
Q = M ΔT = M
(T₀₁ -
)
Q₁ = Q
= T₀₁ - Q / M ce
= 26.0 - 2,664 10⁴ / (0.860 4186)
= 26.0 - 7.40
= 18.6 ° C
The initial temperature of water that has just melted is T₀₂ = 0ª
The initial temperature of the liquid water is T₀₁= 18.6
m
+ M
= M
T₀₁ - m
T₀₂o2
= (M To1 - m To2) / (m + M)
= (0.860 18.6 - 0.080 0) / (0.080 + 0.860)
= 17º C
gg