2 g steam at 100-C is mixed with 5 g ice at -40-C in an ideal calorimeter- The final temperature in -C of the system isLv-500 cal-g- sice-0-5 cal-g-C- sw-1 cal-g-C- Lf- 80 cal-g

We know that: Q=mL, Q=msΔθ Given: m_1=2 g, θ_1=100^∘C, m_2=5 g, θ_2= 40^∘C, L_v=500 cal/g, s_ice=0.5 cal/g^∘C, s_w=1 cal/g^∘C, L_f= 80 cal/g Heat give ...

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Standard VII

Physics

Hot and Cold

2 g steam at ...

Question

$2 g$ steam at $100^{\circ} C$ is mixed with $5 g$ ice at $- 40^{\circ} C$ in an ideal calorimeter. The final temperature (in $^{\circ} C$ ) of the system is
$(L_{v} = 500 c a l / g, s_{i c e} = 0.5 c a l / g^{\circ} C, s_{w} = 1 c a l / g^{\circ} C, L_{f} = 80 c a l / g)$

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Solution

We know that: $Q = m L, Q = m s Δ θ$

Given: $m_{1} = 2 g, θ_{1} = 100^{\circ} C, m_{2} = 5 g, θ_{2} = - 40^{\circ} C, L_{v} = 500 c a l / g, s_{i c e} = 0.5 c a l / g^{\circ} C, s_{w} = 1 c a l / g^{\circ} C, L_{f} = 80 c a l / g$

Heat given by steam in converting into water is

$Q_{1} = m_{1} L_{v}$

$= 2 \times 500$

$= 1000 c a l \dots (i)$

Heat taken by ice to convert to water at $100^{\circ} C$ fully is

$Q_{2} = m_{2} s_{i c e} (0 - (- 40)) + m_{2} L_{f} + m_{2} s_{w} (100 - 0)$

$Q_{2} = (5 \times 40 \times 0.5) + (5 \times 80) + (5 \times 100 \times 1)$

$Q_{2} = 1000 c a l$

As $Q_{1} = Q_{2}$

So, steam and water both are at $100^{\circ} C$ .

Final answer: $100^{\circ} C$ .

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Hot and Cold

Standard VII Physics

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2 g steam at 100∘C is mixed with 5 g ice at −40∘C in an ideal calorimeter. The final temperature (in ∘C) of the system is (Lv=500 cal/g,sice=0.5 cal/g∘C,sw=1 cal/g∘C,Lf=80 cal/g)

$2 g$ steam at $100^{\circ} C$ is mixed with $5 g$ ice at $- 40^{\circ} C$ in an ideal calorimeter. The final temperature (in $^{\circ} C$ ) of the system is
$(L_{v} = 500 c a l / g, s_{i c e} = 0.5 c a l / g^{\circ} C, s_{w} = 1 c a l / g^{\circ} C, L_{f} = 80 c a l / g)$