Thermochemistry Exercises 11

Question

A piece of metal (34.9 g) was heated to 95.0 degrees Celsius and then added to 550-g water having an initial temperature of 28.2 degrees Celsius. If the final temperature of the water and the metal is 31.2 degrees Celsius, calculate the specific heat (s) of the metal. Assume there is no loss of energy to the surroundings! Show work!

Answer 1

The hot metal will lose heat to the cooler water which will gain the heat. This will occur until thermal equilibrium is reached at which point the temperature of the metal and the water will be 31.2ºC

The equation we use for such a reaction is q = mc∆T

q = heat; m = mass; c = specific heat; ∆T = change in temperature (we use c for specific heat, but you can substitute (s) since that is what your question uses as specific heat).

For heat lost by the metal: q_m = m_m x c_m x ∆T_m (you could use -q, but I prefer to keep all values positive)

For heat gained by the water: q_w = m_w x c_w x ∆T_w

Since heat LOST must equal heat GAINED, we then have...

m_m x c_m x ∆T_m = m_w x c_w x ∆T_w

Plugging in the values, and solving for C_m (specific heat of the metal), we have...

(34.9 g)(c_m)(95.0º - 31.2º) = (550 g)(4.184 J/gº)(31.2º -28.2º)

(34.9 g)(c_m)(63.8º) = (550 g)(4.184 J/gº)(3º)

2227c_m = 6904

c_m = 3.10 J/gº = specific heat of the metal