The heat is exchanged when two different temperature objects come in contact. The energy gained by an ice block is 2.3 Joules.
<h3>What is temperature?</h3>
Temperature is the degree of hotness and coldness of the object.
A 7g block of ice was added to a coffee cup full of 103.4 grams of water. The water had an initial temperature T₁ = 24.5 C and a final temperature T₂ = 19.2 C after all the ice had melted.
Heat lost by water = Heat gained by ice
Qgain = ms(T₂ -T₁ )
Substituting the value for mass of water m =103.4 g= 0.1034 kg , specific heat of water s = 4.18 kJ/kg and temperature values, we get
Qgain = 0.1034 x 4.18 x (24.5 - 19.2)
Qgain = 2.3 Joules
Thus, the energy gained by an ice block is 2.3 Joules.
Learn more about temperature.
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Shadows are formed when an opaque object or an object that doesn't allow light to pass through is in the way or infront of etc. a source of light.
Answer: Work can be calculated with the equation: Work = Force × Distance. The SI unit for work is the joule (J), or Newton • meter (N • m). One joule equals the amount of work that is done when 1 N of force moves an object over a distance of 1 m.
Explanation:
Answer:
i 5.3 cm ii. 72 cm
Explanation:
i
We know upthrust on iron = weight of mercury displaced
To balance, the weight of iron = weight of mercury displaced . So
ρ₁V₁g = ρ₂V₂g
ρ₁V₁ = ρ₂V₂ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₂ = density of mercury = 13.6 g/cm³ and V₂ = volume of mercury displaced = ?
V₂ = ρ₁V₁/ρ₂ = 7.2 g/cm³ × 10³ cm³/13.6 g/cm³ = 529.4 cm³
So, the height of iron above the mercury is h = V₂/area of base iron block
= 529.4 cm³/10² cm² = 5.294 cm ≅ 5.3 cm
ρ₁V₁g = ρ₂V₂g
ii
ρ₁V₁ = ρ₃V₃ where ρ₁ = density of iron = 7.2 g/cm³ and V₁ = volume of iron = 10³ cm³ and ρ₃ = density of water = 1 g/cm³ and V₃ = volume of water displaced = ?
V₃ = ρ₁V₁/ρ₃ = 7.2 g/cm³ × 10³ cm³/1 g/cm³ = 7200 cm³
So, the height of column of water is h = V₃/area of base iron block
= 7200 cm³/10² cm² = 72 cm
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