How many degrees will the temperature of four liters of water, which was transferred to the amount of heat equal to 168 kJ, increase? (from water = 1000 kg / m³, from water = 4190 J / kg ° С)
Initial data: V (volume of water) = 4 l = 4 * 10 ^ -3 m ^ 3; Q (amount of heat transferred to water) = 168 kJ = 168 * 10 ^ 3 J.
Reference data: according to the condition ρ (water density) = 1000 kg / m ^ 3; C (specific heat capacity of water) = 4190 J / (kg * K).
The increase in water temperature can be calculated from the formula: Q = C * m * Δt, where m (water mass) = V * ρ.
Q = C * V * ρ * Δt and Δt = Q / (C * V * ρ).
Let’s calculate: Δt = 168 * 10 ^ 3 / (4190 * 4 * 10 ^ -3 * 1000) ≈ 10 ºС.
Answer: The water temperature will increase by 10 ºС.
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