Question

a piece of chromium metal with a mass of 24.26g is heated in boiling water to 98.3 C and then dropped into a coffee-cup calorimeter containing 82.3g of water at 23.3 C. When thermal equilibrium is reached the final temperature is 25.6 C. Calculate the specific heat capacity of chromium

Answer 1

1) This problem can be summarized thusly:

-q_{lost by chromium} = q_{gained by water}

 

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 ^oC^-1 g^-1)

T_f = final temperature (^oC)

T_i = initial temperature (^oC)

 

The specific heat capacity of water is 4.20 J ^oC^-1 g^-1

2) Thus:

q_{lost by chromium} = (24.26 g) × C_chromium × (25.6 - 98.3)°C

q_{lost by chromium} = -1763.702 × C_chromium

 

q_{gained by water} = (82.3 g) × (4.20 J ^oC^-1 g^-1) × (25.6 - 23.3)°C

q_{gained by water} = 795.018 J

3) Therefore,

-q_{lost by chromium} = q_{gained by water}

(1763.702 ^oC g) × C_chromium = (795.018 J)

C_chromium = (795.018 J) / (1763.702 ^oC g) = 0.45 J ^oC^-1 g^-1

C_chromium = 0.45 J ^oC^-1 g^-1

Answer: The specific heat capacity of chromium is 0.45 J ^oC^-1 g^-1