Alab assistant drops a 400.0-g piece of metal at 100.0°c into a 100.0-g aluminum cup containing 500.0 g of water at in a few minutes, she measures the final temperature of the system to be 40.0°c. what is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? the specific heat of this aluminum is 900.0 j/kg ∙ k and that of water is 4186 j/kg ∙ k.
a) 1900 j/kg k
b) 2270 j/kg. k
c) 3300 j/kg k
d) 3800 j/kg k
e) 4280 j/kg k

2274 J/kg ∙ K

Explanation:

The complete statement of the question is :

A lab assistant drops a 400.0-g piece of metal at 100.0°C into a 100.0-g aluminum cup containing 500.0 g of water at 15 °C. In a few minutes, she measures the final temperature of the system to be 40.0°C. What is the specific heat of the 400.0-g piece of metal, assuming that no significant heat is exchanged with the surroundings? The specific heat of this aluminum is 900.0 J/kg ∙ K and that of water is 4186 J/kg ∙ K.

= mass of metal = 400 g

= specific heat of metal = ?

= initial temperature of metal = 100 °C

= mass of aluminum cup = 100 g

= specific heat of aluminum cup = 900.0 J/kg ∙ K

= initial temperature of aluminum cup = 15 °C

= mass of water = 500 g

= specific heat of water = 4186 J/kg ∙ K

= initial temperature of water = 15 °C

= Final equilibrium temperature = 40 °C

Using conservation of energy

heat lost by metal = heat gained by aluminum cup + heat gained by water

answer is c

explanation:

beta is electron from nucleus

something from nucleus has changed, but charge must be same for cons of charge