Q72 / 75 · JEE Main 2026 (05 April Shift 2) · Physics
The heat extracted out of $x$ gram of water initially at $50°C$ to cool it down to $0°C$ is sufficient to evaporate $(1000 - x)$ gram of water also initially at $50°C$. The value of $x$ (closest integer) is _______.
(Take latent heat of water $2256\text{ kJ/kg.K}$, specific heat capacity of water $4200\text{ J/kg.K}$)
(Take latent heat of water $2256\text{ kJ/kg.K}$, specific heat capacity of water $4200\text{ J/kg.K}$)
Correct answer: 922
Solution
Heat extracted to cool $x$ gram of water from $50^{\circ}\text{C}$ to $0^{\circ}\text{C}$ is given by:
$Q_1 = m_1 c \Delta T_1$
$Q_1 = x \times 4.2 \times (50 - 0) = 210x \text{ J}$
Heat required to raise the temperature of $(1000 - x)$ gram of water from $50^{\circ}\text{C}$ to $100^{\circ}\text{C}$ and evaporate it is given by:
$Q_2 = m_2 c \Delta T_2 + m_2 L$
$Q_2 = (1000 - x) \times [4.2 \times (100 - 50) + 2256]$
$Q_2 = (1000 - x) \times (210 + 2256) = 2466(1000 - x) \text{ J}$
Equating the heat extracted and the heat required:
$Q_1 = Q_2$
$210x = 2466(1000 - x)$
$210x = 2466000 - 2466x$
$2676x = 2466000$
$x = \dfrac{2466000}{2676} \approx 921.52$
The closest integer value for $x$ is $922$.
Answer: $922$
$Q_1 = m_1 c \Delta T_1$
$Q_1 = x \times 4.2 \times (50 - 0) = 210x \text{ J}$
Heat required to raise the temperature of $(1000 - x)$ gram of water from $50^{\circ}\text{C}$ to $100^{\circ}\text{C}$ and evaporate it is given by:
$Q_2 = m_2 c \Delta T_2 + m_2 L$
$Q_2 = (1000 - x) \times [4.2 \times (100 - 50) + 2256]$
$Q_2 = (1000 - x) \times (210 + 2256) = 2466(1000 - x) \text{ J}$
Equating the heat extracted and the heat required:
$Q_1 = Q_2$
$210x = 2466(1000 - x)$
$210x = 2466000 - 2466x$
$2676x = 2466000$
$x = \dfrac{2466000}{2676} \approx 921.52$
The closest integer value for $x$ is $922$.
Answer: $922$
Hence, the correct answer is 922.