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Physics

A car's cooling system uses water to absorb heat from the engine. If 5 kg of water absorbs 420 kJ of heat, what is the temperature increase of water? (Specific heat capacity of water = 4200 J kg⁻¹ K⁻¹)

Calorimetry

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Answer

Given,

Mass of water (m) = 5 kg

Heat absorbed (Q) = 420 kJ = 420×103420\times10^3 J

Specific heat capacity of water (c) = 4200 J kg⁻¹ K⁻¹

Let, increase in temperature be △t.

From relation,

Q=mctQ = m c △t

On rearranging terms,

t=Qmc△t = \dfrac{Q}{mc}

Substituting the values in the formula above we get,

t=420×1035×4200=1005=20K=20 oC△t = \dfrac{420\times10^3}{5\times4200}=\dfrac{100}{5}=20 K= 20\ ^oC

Temperature of water will increase by 20 °C.

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