Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
Sign in
Please select an account to continue using cracku.in
↓ →
Join Our JEE Preparation Group
Prep with like-minded aspirants; Get access to free daily tests and study material.
A piece of ice at $$-10^\circ\text{C}$$ is heated to $$-1^\circ\text{C}$$ using a certain quantity of energy. Then another 20 times as much energy is necessary to finally obtain water. Using that the specific heat of ice is half of the specific heat of $$4.2\text{ kJ}/(\text{kg}\,^\circ\text{C})$$ of water, determine the heat of fusion of ice (in kJ/kg) from the above measurement data.
Correct Answer: 375.9
Per kg, heating ice from $$-10^\circ\text{C}$$ to $$-1^\circ\text{C}$$ takes $$2.1 \times 9 = 18.9\text{ kJ}$$. The next stage takes $$20 \times 18.9 = 378\text{ kJ}$$, which warms the ice from $$-1^\circ\text{C}$$ to $$0^\circ\text{C}$$ (2.1 kJ) and then melts it. So the heat of fusion is $$378 - 2.1 = 375.9\text{ kJ/kg}$$.
Was this solution helpful?
Predict your JEE Main percentile, rank & performance in seconds
Educational materials for JEE preparation
Ask our AI anything
AI can make mistakes. Please verify important information.
AI can make mistakes. Please verify important information.