Answer to Question #297268 in Chemistry for Sandra

Question

What is the final temperature when 150.0 g of water at 90.0oC is added to 100.0 g of water at 30.0 oC?

Answer 1

Solution:

This problem can be summarized thusly:

-q_{lost by water1} = q_{gained by water2}

q = m × C × ΔT

q = m × C × (T_f - T_i),

where:

q = amount of heat energy gained or lost by substance (J)

m = mass of sample (g)

C = specific heat capacity (J °C^-1 g^-1)

T_f = final temperature (°C)

T_i = initial temperature (°C)

The specific heat capacity of water is 4.184 J °C^-1 g^-1

 

Therefore:

q_{lost by water1} = (150.0) × (4.184) × (T_f - 90.0)

q_{lost by water1} = 627.6 × (T_f - 90.0)

 

q_{gained by water2} = (100.0) × (4.184) × (T_f - 30.0)

q_{gained by water2} = 418.4 × (T_f - 30.0)

 

-q_{lost by water1} = q_{gained by water2}

-627.6 × (T_f - 90.0) = 418.4 × (T_f - 30.0)

-1.5 × (T_f - 90.0) = T_f - 30.0

-1.5 × T_f + 135 = T_f - 30.0

2.5 × T_f = 165

T_f = 66°C

Answer: The final temperature (T_f) is 66°C