Question

a 1.59 kg1.59 \mathrm{~kg} sample of platinum at 78.2C78.2^{\circ} \mathrm{C} gives off 1.05kcal1.05 \mathrm{kcal} of heat (cF=0.032calg1C1)\left(c_F=0.032 \mathrm{cal} \mathrm{g}^{-1}{ }^{\circ} \mathrm{C}^{-1}\right).

Solution

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Answered 1 year ago
Answered 1 year ago
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b) Given information:

m=1.59kg=1590gTi=78.2Cq=1.05kcal=1050calsp htPt=0.032cal/gC\begin{aligned} m&=1.59\text{kg}=1590\text{g}\\ T_i&=78.2^{\circ}\text{C}\\ q&=-1.05\text{kcal}=-1050\text{cal}\\ sp\ ht_{Pt}&=\mathrm{0.032cal/g^{\circ}C} \end{aligned}

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