Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

The heat of fusion of water is 334 \mathrm{~J} / \mathrm{g}, and its heat of vaporization is 2260 \mathrm{~J} / \mathrm{g}.Calculate the total heat required to convert a 250 \mathrm{~g} ice cube at -10^{\circ} \mathrm{C} into steam at 120^{\circ} \mathrm{C}. Assume no heat is lost to the surroundings and that the specific heat capacity of ice and steam is 2.09 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C}.

Option: 1

160 \mathrm{~J} / \mathrm{K}


Option: 2

1455.19 \mathrm{~J} / \mathrm{K}


Option: 3

200 \mathrm{~J} / \mathrm{K}


Option: 4

20 \mathrm{~J} / \mathrm{K}


Answers (1)

best_answer

To calculate the total heat required, we need to consider the following steps:

1. \text{Heating the ice from }-10^{\circ} \mathrm{C}$ to $0^{\circ} \mathrm{C}$. 2. \text{Melting the ice at }$0^{\circ} \mathrm{C}$. 3. \text{Heating the water from }$0^{\circ} \mathrm{C}$ to $100^{\circ} \mathrm{C}$. 4. \text{Vaporizing the water at} $100^{\circ} \mathrm{C}$. 5. \text{Heating the steam from} $100^{\circ} \mathrm{C}$ to $120^{\circ} \mathrm{C}.
Let's calculate the heat for each step:

\begin{aligned} Q_{1} & =m \cdot c \cdot \Delta T \\ & =250 \mathrm{~g} \cdot 2.09 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C} \cdot(0-(-10))^{\circ} \mathrm{C} \\ & =5225 \mathrm{~J} \end{aligned}

\begin{aligned} Q_{2} & =m \cdot \text { Heat of Fusion } \\ & =250 \mathrm{~g} \cdot 334 \mathrm{~J} / \mathrm{g} \\ & =83500 \mathrm{~J} \end{aligned}

\begin{aligned} & Q_{3}=m \cdot c \cdot \Delta T \\ & =250 \mathrm{~g} \cdot 2.09 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C} \cdot(100-0)^{\circ} \mathrm{C} \\ & =52250 \mathrm{~J} \\ & Q_{4}=m \cdot \text { Heat of Vaporization } \\ & =250 \mathrm{~g} \cdot 2260 \mathrm{~J} / \mathrm{g} \\ & =565000 \mathrm{~J} \\ & Q_{5}=m \cdot c \cdot \Delta T \\ & =250 \mathrm{~g} \cdot 2.09 \mathrm{~J} / \mathrm{g}^{\circ} \mathrm{C} \cdot(120-100)^{\circ} \mathrm{C} \\ & =5225 \mathrm{~J} \end{aligned}

The total heat required is the sum of these values:
\begin{aligned} Q_{\text {total }} & =Q_{1}+Q_{2}+Q_{3}+Q_{4}+Q_{5} \\ & =5225 \mathrm{~J}+83500 \mathrm{~J}+52250 \mathrm{~J}+565000 \mathrm{~J}+5225 \mathrm{~J} \\ & =687200 \mathrm{~J} \end{aligned}

Therefore, the total heat required is 687200 \mathrm{~J}.
Therefore, the correct option is (B).

Posted by

Gautam harsolia

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2026 application correction begins on jeemain.nta.nic.in; edit details by December 2
anam-khan

JEE Mains 2026 LIVE: NTA to conduct session 1 from January 21 to 30; exam pattern, syllabus
anam-khan

JEE Main 2026 correction window opens tomorrow; NTA allows exam city change

Latest Question