Calorimetry And Mixtures

Question

A 200 g iron block at $150^\circ	ext{C}$ is dropped into 400 g of water at $20^\circ	ext{C}$ in an insulated container. Taking the specific heat of iron as $0.45\,	ext{J g}^{-1}\,^\circ	ext{C}^{-1}$ and of water as $4.2\,	ext{J g}^{-1}\,^\circ	ext{C}^{-1}$, what is the approximate final equilibrium temperature?

RankGuru · Accepted Answer

Calorimetry rests on conservation of energy in an insulated system: the heat lost by the hotter body equals the heat gained by the cooler body, with no exchange to the surroundings. Let the equilibrium temperature be $T$. Heat lost by iron is $m_i c_i(150 - T)$ and heat gained by water is $m_w c_w(T - 20)$. Setting them equal, $200	imes0.45	imes(150 - T) = 400	imes4.2	imes(T - 20)$, which gives $90(150 - T) = 1680(T - 20)$. Expanding, $13500 - 90T = 1680T - 33600$, so $47100 = 1770T$ and $T \approx 26.6^\circ	ext{C}$, closest to $26^\circ	ext{C}$. The value $32^\circ	ext{C}$ would follow from underestimating the water's large heat capacity. The values $45^\circ	ext{C}$ and $58^\circ	ext{C}$ wrongly let the small iron mass dominate the mixture. This is the standard NCERT method of mixtures using the principle of calorimetry, valid here because the container is insulated and no heat leaks to the surroundings or is absorbed by the vessel walls. The product of mass and specific heat, called the thermal capacity, decides how strongly each body resists temperature change, and the water's capacity here vastly exceeds the iron's. As a plausibility check, water has a far higher specific heat and greater mass, so the final temperature should sit very close to the water's initial $20^\circ	ext{C}$, exactly as the modest rise to about $26^\circ	ext{C}$ confirms.

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