Calorimetry And Heat Capacity

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How much heat energy must be supplied to raise the temperature of 0.5 kg of water from 25°C to 75°C, given that the specific heat of water is 4200 J/kg°C?

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NCERT Class 11, Chapter 11 (Thermal Properties of Matter) defines the heat required to change a substance's temperature as $Q = mc\Delta T$, where $m$ is mass, $c$ the specific heat capacity, and $\Delta T$ the temperature change. Specific heat measures how much energy one kilogram of a substance needs to warm by one degree. Here $\Delta T = 75 - 25 = 50\ ^{\circ}	ext{C}$. Substituting $m = 0.5\ 	ext{kg}$, $c = 4200\ 	ext{J/kg}^{\circ}	ext{C}$ and $\Delta T = 50$ gives $Q = 0.5 	imes 4200 	imes 50 = 105000\ 	ext{J}$. The option 52500 J halves the temperature rise wrongly. The option 210000 J ignores the half-kilogram mass. The option 21000 J uses a 10°C rise instead of 50°C. As a magnitude check, water's high specific heat means even modest heating demands a large amount of energy, which is why water is an effective coolant, confirming that 105 kJ for this process is reasonable in both value and joule units.

Calorimetry And Heat Capacity

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