Find the temperature

3 liquids which can't react with each other chemically are mixed together in a calorimeter.

What is the temperature of all 3 liquids when they are in heat equilibrium? Type your answer in Celcius degree .

Note: Assume that there is no heat escaping from the calorimeter and no heat flow between the calorimeter and the liquids. Moreover, no liquids turn solid or into gases.

The mass, specific heat capacity and temperature of each liquid is given in the table below:

Liquids # 1 1 # 2 2 # 3 3
Mass ( k g kg ) 1 1 10 10 5 5
Specific heat capacity ( J / k g . K J/kg.K ) 2000 2000 4000 4000 2000 2000
Temperature ( o C ^oC ) 6 6 40 -40 60 60


The answer is -19.

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2 solutions

Steven Chase
Jun 11, 2020

Let the masses be m 1 , m 2 , m 3 m_1, m_2, m_3 . Let the specific heats be C 1 , C 2 , C 3 C_1, C_2, C_3 . Let the initial temperatures be T 1 , T 2 , T 3 T_1, T_2, T_3 , and the final temperature be T T . Determine a value of T T such that the net gain/loss of energy is zero.

m 1 C 1 ( T T 1 ) + m 2 C 2 ( T T 2 ) + m 3 C 3 ( T T 3 ) = 0 m_1 C_1 (T - T_1) + m_2 C_2 (T - T_2) + m_3 C_3 (T - T_3) = 0

Solving for T T yields T = 19 T = -19

The final temperature is

1 × 2000 × 6 + 10 × 4000 × ( 40 ) + 5 × 2000 × 60 1 × 2000 + 10 × 4000 + 5 × 2000 = 19 ° \dfrac{1\times 2000\times 6+10\times 4000\times (-40)+5\times 2000\times 60}{1\times 2000+10\times 4000+5\times 2000}=\boxed {-19}\degree Celsius.

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