Answer to Question #225023 in Chemistry for marlou

Question

A piece of titanium metal with a mass of 20.8 g is

heated in boiling water to 99.5°C and then dropped

into a coffee cup calorimeter containing 75.0 g of

water at 21.7°C. When thermal equilibrium is reached,

the final temperature is 24.3°C. Calculate the specific

heat capacity of titanium.

Answer 1

Solution:

1) This problem can be summarized thusly:

-q_{lost by titanium} = 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

m = mass of sample

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

T_f = final temperature

T_i = initial temperature

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

 

2) Therefore:

q_{lost by titanium} = (20.8 g) × C_titanium × (24.3 - 99.5)°C

q_{lost by titanium} = -1564.16 × C_titanium

 

q_{gained by water} = (75.0 g) × (4.20 J g^-1 °C^-1) × (24.3 - 21.7)°C

q_{gained by water} = 819

 

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

-(-1564.16 × C_titanium) = 819

C_titanium = (819) / (1564.16) = 0.5236 = 0.524

C_titanium = 0.524 J g^-1 °C^-1

Answer: The specific heat capacity of titanium is 0.524 J g^-1 °C^-1