Question

35.5 grams of an unknown substance is heated to 103.0 degrees Celsius and then placed into a calorimeter containing 100.0 grams of water at 24.0 degrees Celsius. If the final temperature reached in the calorimeter is 29.5 degrees Celsius, what is the specific heat of the unknown substance?

Answer 1

E = m × c × θ
E is the energy transferred in joules, J
m is the mass of the substances in kg
c is the specific heat capacity in J / kg °C
θ (‘theta’) is the temperature change in degrees Celsius, °C
θ=(Final temperature-Initial temperature)
θ=29.5-24.0=5.5°C
E(unknown substance)=E of water
35.5*x*(29.5-103)=100*4.18*5.5

Answer 2

thank you

Answer 3

your welcome :)

Answer 4

so the specific heat of the unknown substance is -0.88??

Answer 5

since the change in temperature of the unknown is negative, the heat energy released will also be negative. The specific heat capacity is always a positive value. 0.88 may be the right answer, but it will be positive