One flask contains water, the other contains kerosene of the same mass and temperature.

One flask contains water, the other contains kerosene of the same mass and temperature. One flask contains water, the other contains kerosene of the same mass and temperature. An equally heated iron cube was thrown into each flask. Which will heat up to a higher temperature – water or kerosene?

One flask contains water with heat capacity c₁ = 4200 J / (kg ∙ ° C), and in the other – kerosene with heat capacity c₂ = 2100 J / (kg ∙ ° C) of the same mass m₁ and temperature t₀. An equally heated iron cube of mass m₂, temperature t and heat capacity c = 460 J / (kg ∙ ° C) was thrown into each flask. In this case, the water will be heated to the temperature t₁, and the kerosene will be heated to the temperature t₂. The heat balance equation for water will be:

m ∙ s₁ ∙ (t₁ – t₀) = m₂ ∙ s ∙ (t – t₁), then t₁ = (m₂ ∙ s ∙ t + m ∙ s₁ ∙ t₀) / (m ∙ s₁ + m₂ ∙ s) or t₁ = ( m₂ ∙ 460 ∙ t + m ∙ 4200 ∙ t₀) / (m ∙ 4200 + m₂ ∙ 460).

And for kerosene:

m ∙ s₂ ∙ (t₂ – t₀) = m₂ ∙ s ∙ (t – t₂), then t₂ = (m₂ ∙ s ∙ t + m ∙ s₂ ∙ t₀) / (m ∙ s₂ + m₂ ∙ s) or t₂ = ( m₂ ∙ 460 ∙ t + m ∙ 2100 ∙ t₀) / (m ∙ 2100 + m₂ ∙ 460).

Since t₁ – t₂ <0, then t₁ <t₂.

Answer: kerosene will be heated to a higher temperature.