5 text{ g} of steam at 100^{circ} text{C} is | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given: Mass of steam,$m_s = 5 	ext{ g}$,Temperature of steam,$T_s = 100^{\circ} 	ext{C}$.Mass of ice,$m_i = 5 	ext{ g}$,Temperature of ice,$T_i

Vedclass · Accepted Answer

$100$

Vedclass · Answer

(A) Given: Mass of steam,$m_s = 5 	ext{ g}$,Temperature of steam,$T_s = 100^{\circ} 	ext{C}$.Mass of ice,$m_i = 5 	ext{ g}$,Temperature of ice,$T_i = 0^{\circ} 	ext{C}$.Latent heat of vaporization,$L_v = 540 	ext{ cal/g}$.Latent heat of fusion,$L_f = 80 	ext{ cal/g}$.Specific heat of water,$s = 1 	ext{ cal/g}^{\circ} 	ext{C}$.Step $1$: Calculate heat released by steam to condense into water at $100^{\circ} 	ext{C}$:$Q_1 = m_s 	imes L_v = 5 	imes 540 = 2700 	ext{ cal}$.Step $2$: Calculate heat required by ice to melt and reach $100^{\circ} 	ext{C}$:Heat to melt ice: $Q_{melt} = m_i 	imes L_f = 5 	imes 80 = 400 	ext{ cal}$.Heat to raise water temperature from $0^{\circ} 	ext{C}$ to $100^{\circ} 	ext{C}$: $Q_{rise} = m_i 	imes s 	imes \Delta T = 5 	imes 1 	imes 100 = 500 	ext{ cal}$.Total heat required by ice: $Q_2 = 400 + 500 = 900 	ext{ cal}$.Step $3$: Compare $Q_1$ and $Q_2$:Since $Q_1 > Q_2$,the steam has more than enough energy to melt the ice and heat the resulting water to $100^{\circ} 	ext{C}$.Therefore,the final mixture will be at $100^{\circ} 	ext{C}$ with some steam remaining.

$5 \text{ g}$ of steam at $100^{\circ} \text{C}$ is mixed with $5 \text{ g}$ of ice at $0^{\circ} \text{C}$. What is the final temperature of the mixture (in $^{\circ} \text{C}$)?

Similar Questions

The amount of heat required to change $1 \, g$ of ice at $0^\circ C$ into water at $100^\circ C$ is ............ $cal$.

Two different metal wires are joined to form a thermocouple. One junction is kept at a constant temperature,while the other junction is heated to a higher temperature. Which graph represents the variation of thermo $e.m.f.$ $(E)$ with temperature $(t)$?

If there are no heat losses,the heat released by the condensation of $x \, g$ of steam at $100^\circ C$ into water at $100^\circ C$ can be used to convert $y \, g$ of ice at $0^\circ C$ into water at $100^\circ C$. Then the ratio $y : x$ is nearly (in $:1$)

Work done in converting $1 \, g$ of ice at $-10^{\circ} C$ into steam at $100^{\circ} C$ is .......... $kJ$.

The freezing point of a liquid decreases when pressure is increased,if the liquid:

Vedclass Test Series

Exam Paper Generator

Online Exam Module