Answer:
The volume increases because the temperature increases and is 2.98L
Explanation:
Charles's law states that the volume of a gas is directely proportional to its temperature. That means if a gas is heated, its volume will increase and vice versa. The equation is:
V₁/T₁ = V₂/T₂
<em>Where V is volume and T is absolute temperature of 1, initial state, and 2, final state of the gas.</em>
In the problem, the gas is heated, from 53.00°C (53.00 + 273.15 = 326.15K) to 139.00°C (139.00 + 273.15 = 412.15K).
Replacing in the Charles's law equation:
2.36L / 326.15K= V₂/412.15K
<h3>2.98L = V₂</h3>
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DR is the correct answer
hope this helps
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Note: The question is incomplete. The complete question is given below :
Suppose a substance has a heat of fusion equal to 45 cal/g and a specific heat of 0.75 cal/g°C in the liquid state. If 5.0 kcal of heat are applied to a 50 g sample of the substance at a temperature of 24°C, what will its new temperate be? What state will the sample be in? (melting point of the substance = 27°C; specific heat of the solid =0.48 cal/g°C; boiling point of the substance = 700°C)
Explanation:
1.a) Heat energy required to raise the temperature of the substance to its melting point, H = mcΔT
Mass of solid sample = 50 g; specific heat of solid = 0.75 cal/g; ΔT = 27 - 24 = 3 °C
H = 50 × 0.75 × 3 = 112.5 calories
b) Heat energy required to convert the solid to liquid at its melting point at 27°C, H = m×l, where l = 45 cal/g
H = 50 × 45 = 2250 cal
c) Total energy used so far = 112.5 cal + 2250 cal = 2362.5 calories.
Amount of energy left = 5000 - 2362.5 = 2637.5 cal
The remaining energy is used to heat the liquid
H = mcΔT
Where specific heat of the liquid, c = 0.75 cal/g/°C, H = 2637.5 cal, ΔT = temperature change
2637.5 = 50 × 0.75 x ΔT
ΔT = 2637.5 / ( 50*0.75)
ΔT = 70.3 °C
Final temperature of sample = (70.3 + 27) °C = 97.3 °C
The substance will be in liquid state at a temperature of 97.3 °C
i hope that this eg gonna help u
