Question

$1 kg$ ice at $-10°C$ is mixed with $1 kg$ of water at $100°C$. Find the final equilibrium temperature and what will be the mixture content?

Answer 1

According to the question we have ice and water. When they both are mixed we get a mixture of ice water at a different temperature. We will assume the final temperature as T and by applying the principle of calorimetry which states the law of conservation energy, i.e. the total heat lost by the hot body is equal to the total heat gained by the cold body.

Complete step by step answer:
We know specific heat of water =1 cal /gm/°C, Specific heat of ice =0.5 cal/gm/°C, Latent heat of fusion =80cal/gm
We assume the equilibrium temperature for the mixture is at T. When we put ice into water at 100°C it will lose heat and ice will absorb heat released by water at 100°C.

For the mixture heat released =heat absorbed
Mass of water $\times$ specific heat of water = (mass of ice × specific heat of ice) + (weight of ice that turned into water ×latent heat of fusion) + (mass of water ×specific heat of water )
$100 - T = \left( {1kg \times 0.5cal/gm/^\circ C} \right) \times 10 + \left[ {\left( {1kg \times 80cal/gm} \right) + \left( {1kg \times 1cal/gm/^\circ C} \right)} \right]T$
$\Rightarrow 100 - T = 5 + 80 + T$
$\Rightarrow 2T = 15$
$\therefore T = 7.5^\circ C$

Now the second part of the question asks about the content of the mixture, to which we can simply answer the mixture is content as $2kg$ of water at $7.5°C$. This is because the mixture will have both ice converted to water and water present and the weight of the two quantities is $1 kg$ each, so their sum will be equal to $2 kg$.

Hence,equilibrium temperature is $7.5°C$ and mixture content is $2 kg$.

Note: While answering these types of questions pay attention to the three phases of ice which it goes through to turn into water. First ice absorbs heat and it melts then the melting ice has latent heat of fusion then into water state.