Respuesta :
Answer:
The amount of kilograms of ice at -20.0°C that must be dropped into the water to make the final temperature of the system 40.0°C = 0.0674 kg
Explanation:
Heat gained by ice in taking the total temperature to 40°C = Heat lost by the water
Total Heat gained by ice = Heat used by ice to move from -20°C to 0°C + Heat used to melt at 0°C + Heat used to reach 40°C from 0°C
To do this, we require the specific heat capacity of ice, latent heat of ice and the specific heat capacity of water. All will be obtained from literature.
Specific heat capacity of ice = Cᵢ = 2108 J/kg.°C
Latent heat of ice = L = 334000 J/kg
Specific heat capacity of water = C = 4186 J/kg.°C
Heat gained by ice in taking the total temperature to 40°C = mCᵢ ΔT + mL + mC ΔT = m(2108)(0 - (-20)) + m(334000) + m(4186)(40 - 0) = 42160m + 334000m + 167440m = 543600 m
Heat lost by water = mC ΔT = 0.25 (4186)(75 - 40) = 36627.5 J
543600 m = 36627.5
m = 0.0674 kg = 67.4 g of ice.
The mass of the ice that must be dropped into the water to yield the resulting temperature is 0.0874 kg.
The given parameters;
- mass of the water, = 0.25 kg = 250 g
- temperature of the warm water, = 75 ⁰C
- temperature of the ice, = -20 ⁰C
- final temperature of the mixture, = 40 ⁰C
- heat of fusion of ice, = 333.55 J/g
Let the mass of the ice = mₓ
Apply the principle of conservation of energy to determine the mass of the ice;
heat lost by warm water = heat to raise ice to 0⁰C + heat of fusion of ice + heat to raise liquid ice to 40⁰C.
[tex]250 (4.184)(75 - 40) = 4.184m_x(0 --20) \ + \ 333.55m_x \ + \ 4.184m_x(40-0)\\\\250(4.184)(35) = 4.184m_x(20) + 333.55m_x + 4.184(40)\\\\divide \ through \ by \ 4.184\\\\250(35) = 20m_x + 79.72m_x + 40\\\\8710 = 99.72m_x\\\\m_x = \frac{8710}{99.72} \\\\m_x = 87.35 \ g\\\\m_x = 0.0874 \ kg[/tex]
Thus, the mass of the ice that must be dropped into the water to yield the resulting temperature is 0.0874 kg.
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