A can of water of volume `0.5 m^(3)` at a temperature `30^(@)C` is cooled to `15^(@)C`. Neglect heat capacity of the can. If the amount of heat released by water is used to a box of weight 10 kg, find the height to whilch it can be lifted. Given that specific theat of water is 4200`J//kg.^(@)C`. Take `g=10m//s^(2)`.

Answer

To solve the problem, we need to follow these steps:

### Step 1: Calculate the mass of water
Given the volume of water \( V = 0.5 \, m^3 \) and knowing that the density of water \( \rho \) is approximately \( 1000 \, kg/m^3 \):

\[ \text{Mass of water} (m) = \rho \times V = 1000 \, kg/m^3 \times 0.5 \, m^3 = 500 \, kg \]

### Step 2: Calculate the heat released by the water
The specific heat of water \( c = 4200 \, J/(kg \cdot °C) \) and the change in temperature \( \Delta T = T_f - T_i = 15°C - 30°C = -15°C \).

Using the formula for heat transfer:

\[ Q = m \cdot c \cdot \Delta T \]

Substituting the values:

\[ Q = 500 \, kg \cdot 4200 \, J/(kg \cdot °C) \cdot (-15°C) \]
\[ Q = 500 \cdot 4200 \cdot (-15) = -31500000 \, J \]

Since we are interested in the amount of heat released (which is a positive value), we take:

\[ Q = 31500000 \, J \]

### Step 3: Calculate the height to which the box can be lifted
The work done to lift the box is given by the equation:

\[ W = m \cdot g \cdot h \]

Where:
- \( W \) is the work done (or heat released, \( Q \)),
- \( m = 10 \, kg \) (weight of the box),
- \( g = 10 \, m/s^2 \) (acceleration due to gravity),
- \( h \) is the height.

Setting the heat released equal to the work done:

\[ Q = m \cdot g \cdot h \]
\[ 31500000 \, J = 10 \, kg \cdot 10 \, m/s^2 \cdot h \]
\[ 31500000 = 100 \cdot h \]
\[ h = \frac{31500000}{100} = 315000 \, m \]

### Final Answer:
The height to which the box can be lifted is \( 315000 \, m \).

---