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Question

10 g of water at 70°C is mixed with 5 g of water at 30°C. Find the temperature of the mixture in equilibrium.

Moderate

Solution

Let t° C be the temperature of the mixture. From energy conservation,

Heat given by 10 g of water = Heat taken by 5 g of water

$\begin{array}{l}\text{or}{\mathrm{m}}_{1}{\mathrm{C}}_{\text{water}}\left|{\mathrm{\Delta t}}_{1}\right|={\mathrm{m}}_{2}{\mathrm{C}}_{\text{water}}\left|{\mathrm{\Delta t}}_{2}\right|\\ \therefore \left(10\right)(70-\mathrm{t})=5(\mathrm{t}-30)\\ \therefore \mathrm{t}=36{.67}^{\circ}\mathrm{C}\end{array}$

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A bullet of mass 5g , Travelling with a speed of 210 m/s , strikes a fixed wooden target . One half its kinetic energy is converted in to heat in the bullet while the other half is converted in to heat in the wood. The rise of temperature of the bullet if the specific heat of the material is 0.030 cal/$\left({\text{g-}}^{\text{0}}\text{C}\right)\left(1cal=4.2\times {10}^{7}\text{ergs}\right)$ close to:

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