Answer to Question #276770 in Chemistry for michelle

Question

100 g of a liquid at 25 °C is heated to raise its temperature to 50.0 °C. The thermal energy gained by the liquid is 6100 J. The specific heat capacity of the liquid is:

1. 2.44 J/g°C
2. 2.06 J/g°C
3. 2.02 J/g°C
4. 2.00 J/g°C
5. 0.82 J/g°C

Answer 1

Solution:

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 °C^-1 g^-1)

T_f = final temperature (°C)

T_i = initial temperature (°C)

 

Thus:

(6100 J) = (100 g) × C × (50.0°C - 25.0°C)

С = (6100 J) / (100 g × 25.0°C) = 2.44 J °C^-1 g^-1

Therefore, the specific heat capacity of the liquid is 2.44 J °C^-1 g^-1

Answer: (1) 2.44 J/g°C