These tasks: V (volume of heated water) = 4 l (in SI V = 0.004 m3); Q (the amount of heat that the water received) = 168 kJ (168 * 10 ^ 3 J).
Reference values: Sv (specific heat capacity of water) = 4200 J / (kg * K); ρw (density of heated water) = 1000 kg / m3.
The increase in temperature of 4 liters of water can be expressed from the formula: Q = Sv * ρw * V * Δt, whence Δt = Q / (Sv * ρw * V).
Calculation: Δt = 168 * 10 ^ 3 / (4200 * 1000 * 0.004) = 10 ºС.
Answer: The temperature of 4 liters of water will rise by 10 ºС.
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