Can someone please answer this question?

Question

A 27.44 g sample of a substance is initially at 21.0 °C. After absorbing 2565 J of heat, the temperature of the substance is 156.6 °C. What is the specific heat (c) of the substance?

  

C= j/g °C

Answer 1

This a thermochemistry specific heat problem given a substance with a mass of 27.44 grams whose initial temperature is 21 degrees celsius. It absorbs 2565J of energy making this an endothermic process as evidenced by the final temperature being raised to 156.6 degrees celsius. To find the specific heat you need to use this equation:

q = (c)(m)(deltaT)

q = the heat absorbed or released in a process in Joules

C= the specific heat in J/g.c

m= the mass in grams

delta T= final temperature - initial temperature in celsius. When using this formula you need to use celsius not Kelvin as the temperature scale.. In fact if you see a formula that asks for Delta T always use celsius.

inputting the known data you can solve for specific heat,

+ 2565J = (C J/g.c)(27.44 gms)(156.6 degrees C - 21 degrees C) (note the sign for joules is + indicating that heat is absorbed, an endothermic process.

2565J = (C J/g.c)(3720.86)

C = 0.69 J/g.c

Of note the higher the specific heat the more energy is required to raise the temperature (average kinetic energy of a substance) For example, the specific heat of water is 4.184 J/g.c, quite high meaning it takes a substantial amount of Joules of energy to raise the temperature of water.

I hope this was helpful.

Answer 2

Use the basic equation Q = c x m x (t_f - t_i ) where Q is the quantity of heat in joules, c the specific heat, m the mass in grams, and t_f and t_i are are tge final and initial temperatures respectively. Solve for c.

c = Q / (m x (t_f - t_i )) = 2565 j / (27.44g x (156.6 - 21.0)C) = 0.689 j / g °C.