Determine the final temperature after 52.5 g of copper (c = 0.385 j/gc) at 93.6 c is placed into 225 g of water at 21.3 c.

22.82°C

Concept tested: Quantity of heat

We are given;

We are required to calculate the final temperature of the mixture.

Q = m × c × ΔT

Since the final temperature is X°C then the change in temperature, ΔT will be (93.6 - X)°C

Quantity of heat released by copper, Q will be;

  = 52.5 g × 0.385 J/g°C × (93.6 - X)°C

  =  1891.89 - 20.2125X joules

Specific heat capacity of water = 4.184 J/g°C

Change in temperature = (X-21.3)°C

Therefore;

Q = 225 g × 4.184 J/g°C × (X-21.3)°C

   = 941.4X-20051.82 Joules

We know that the quantity of heat released is equivalent to the amount of heat absorbed.

Therefore;

1891.89 - 20.2125X Joules = 941.4X-20051.82 Joules

961.6125X = 21943.71

              X = 22.82°C

Therefore, the final temperature is 22.82°C

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.