On a winter day temperature of the tap water is `20^(@)C` where as the room temperature is `5^(@)C` .Water is stored in a tank of capacity `0.5 m^(3)` for household use .If it were possible to use the heat liberated by the water to lift a `10kg` mass vertically , how high can it be lifted as the water comes to the room temperature ? Take `g = 10ms^(-2)`

Explanation

To solve the problem step by step, we will calculate the heat released by the water as it cools down from `20°C` to `5°C`, and then use that heat to determine how high a `10 kg` mass can be lifted.

### Step 1: Calculate the mass of the water
The mass of the water can be calculated using the formula:
\[ m = V \times \rho \]
where:
- \( V \) is the volume of water (given as `0.5 m³`)
- \( \rho \) is the density of water (approximately `1000 kg/m³`)

Substituting the values:
\[ m = 0.5 \, m³ \times 1000 \, kg/m³ = 500 \, kg \]

### Step 2: Calculate the heat lost by the water
The heat lost by the water when it cools down can be calculated using the formula:
\[ Q = m \times c \times \Delta T \]
where:
- \( Q \) is the heat lost
- \( m \) is the mass of the water (calculated in Step 1)
- \( c \) is the specific heat capacity of water (approximately `4200 J/(kg·K)`)
- \( \Delta T \) is the change in temperature

The change in temperature \( \Delta T \) is:
\[ \Delta T = T_{\text{initial}} - T_{\text{final}} = 20°C - 5°C = 15°C \]

Now substituting the values into the heat loss formula:
\[ Q = 500 \, kg \times 4200 \, J/(kg·K) \times 15 \, K \]
Calculating:
\[ Q = 500 \times 4200 \times 15 = 3.15 \times 10^7 \, J \]

### Step 3: Calculate the height to which the mass can be lifted
The potential energy gained by the mass when lifted is given by:
\[ PE = m \times g \times h \]
where:
- \( PE \) is the potential energy
- \( m \) is the mass being lifted (`10 kg`)
- \( g \) is the acceleration due to gravity (`10 m/s²`)
- \( h \) is the height

We can equate the heat lost by the water to the potential energy gained by the mass:
\[ Q = m \times g \times h \]
Substituting the known values:
\[ 3.15 \times 10^7 \, J = 10 \, kg \times 10 \, m/s² \times h \]

This simplifies to:
\[ 3.15 \times 10^7 = 100 \times h \]

Now solving for \( h \):
\[ h = \frac{3.15 \times 10^7}{100} = 3.15 \times 10^5 \, m \]

### Final Answer
The height to which the `10 kg` mass can be lifted is:
\[ h = 315000 \, m \text{ or } 315 \, km \]