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

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