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

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.  

what is the question because that is not a question.

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )