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URGENTTT PLEASE HELPPPP. You put m1 = 1 kg of ice cooled to -20°C into mass m2 = 1 kg of water at 2°C. Both are in a thermally insulated chamber. For water L = 3.33 x 105 J/kg. The specific heat of ice is 2090 J/(kg°C) and of water 4186 J/(kg°C). How much does the ice heat up in order to bring the water down to 0°C?


A. 0.04°C


B. 0.4°C


C. 4°C


D. 10°C


E. 20°C

Respuesta :

Answer:

Explanation:

heat lost by water will be used to increase the temperature of  ice

heat gained by ice

= mass x specific heat  x rise in temperature

1 x 2090 x t

heat lost by water in cooling to 0° C

= mcΔt  where m is mass of water , s is specific heat of water and Δt is fall in temperature .

= 1 x 2 x 4186  

8372

heat lost = heat gained

1 x 2090 x t  = 8372

t = 4°C

There will be a rise of  4 degree in the temperature of ice.  

 

onyebuchinnaji

The increase in the temperature of the ice to bring the water to 0 °C is 4 ⁰C.

The given parameters;

  • mass of the ice, m₁ = 1 kg
  • temperature of the ice, t₁ = -20°C
  • mass of the water, m₂ =  1 kg
  • temperature of the water, t₂ = 2 °C

Apply the principle of conservation of energy to determine the increase in the temperature of the ice to bring the water to 0 °C.

Heat absorbed by the ice = Heat lost by water

[tex]Q_{ice} = Q_{w}\\\\mc\Delta t_{ice} = mc \Delta t_{w}\\\\1 \times 2090 \times \Delta t = 1 \times 4186 \times (2-0)\\\\2090\Delta t = 8372\\\\\Delta t = \frac{8372}{2090} \\\\\Delta t = 4 \ ^0C[/tex]

Thus, the increase in the temperature of the ice to bring the water to 0 °C is 4 ⁰C.

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