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Question

When (8.140x10^2) J of heat is added to a liquid sample that is initially at 25.0 degrees Celsius, the temperature rises to (2.80x10^1) degrees Celsius. What is the volume of the sample in milliliters? The specific heat of the liquid 0.862 J/(g oC). Density of the liquid is (4.7100x10^0) g/mL.

A (4.5200x10^1) grams of a metal at (1.2100x10^2) degrees Celsius is dropped into (1.14x10^2) grams of water at 15.6 degrees Celsius. When the system establishes an equilibrium, the temperature of the system is (2.6000x10^1) degrees Celsius. What is the specific heat of metal in J/(g oC)? Specific heat of water is 4.184 J/(g oC).  

Answer 1

Use q = mC∆T

q = heat = 8.140x10² J = 814.0 J

m = mass = ?

C = specific heat = 0.862 J/gº

∆T = change in temperature = 2.80x10¹º - 25.0º = 3º

Solving for mass:

m = q / C∆T = 814.0 J / (0862 J/gº)(3º) = 314.8 g

Use the density for find volume:

314.8 g x 1 ml / 4.710 g = 66.8 mls

Final temp = 26.00º

∆T_metal = 121.0 º - 26º = 95º

∆T_water = 26º - 15.6º = 10.4º

heat lost by metal = heat gained by water

(45.20g)(C)(95º) = (114 g)(4.184 J/gº)(10.4º)

Solve for C:

C = (114 g)(4.184 J/gº)(10.4º) / (45.20 g)(95º)

C = 1.16 J/gº