Answer to Question #170404 in Chemistry for Ava Brooke

Question

A 12.40 gram sample is heated to an initial temperature of 50 ^oC. The sample is then placed in a calorimeter containing 6 grams of water starting at 24 ^oC. If the final temperature is 35 ^oC what is the specific heat of the sample?

s_H2O = 4.18

Answer 1

Solution:

1) This problem can be summarized thusly:

q_{lost by sample} = 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 (J)

m = mass of sample (g)

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

T_f = final temperature (^oC)

T_i = initial temperature (^oC)

2) Therefore:

q_sample = (12.40) × C_sample × (35 - 50) = -186 × C_sample

q_water = (6.00 g) × (4.18) × (35 - 24) = 275.88

 

275.88 = 186 × C_sample

C_sample = 1.483 J ^oC^-1 g^-1

Answer: The specific heat of the sample is 1.483 J ^oC^-1 g^-1.