Question

A 100 g sample of an unknown liquid absorbs 2000 j of heat energy, raising the liquid's temperature from 50 ◦ c to 70 ◦

Answer 1

Since there is no phase change, we can use the heat equation,
Q = mcΔT 
where Q is the amount of energy transferred (J), m is the mass of the substance (kg), c is the specific heat (J kg⁻¹ °C⁻¹) and ΔT is the temperature difference (°C).

Q = 2000 J
m = 100 g = 0.1 kg
c = ?
ΔT = (70 °C - 50 °C) = 20 °C

By applying the formula,
2000 J = 0.1 kg x c x 20 °C
         c = 2000 J / (0.1 kg x 20 °C)
         c = 1000 J kg⁻¹ °C⁻¹

Hence, the specific heat capacity of the liquid is 1000 J kg⁻¹ °C⁻¹.