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AleksandrR [38]
3 years ago
15

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 i

nsulated 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
Physics
1 answer:
STatiana [176]3 years ago
5 0

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.  

 

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