10,000J will raise the temperature of 1 liter of water by 2.38 °C
Using Q = mcΔT where Q = quantity of heat supplied = 10000 J,
m = mass of 1 liter of water = ρV where ρ = density of water = 1000 kg/m³ and V = volume of water = 1 liter = 1 dm³ = 1 × 10⁻³ m³.
So, m = ρV = 1000 kg/m³ × 1 × 10⁻³ m³ = 1 kg,
c = specific heat capacity of water = 4200 J/kg°C and ΔT = temperature change.
Making ΔT subject of the formula, we have
ΔT = Q/mc
Substituting the values of the variables into the equation, we have
ΔT = Q/mc
ΔT = 10000 J/(1 kg × 4200 J/kg°C)
ΔT = 10000 J/(4200 J/°C)
ΔT = 2.38 °C
So, 10,000J will raise the temperature of 1 liter of water by 2.38 °C
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