Question

2. 20 g of ice at 0°C is placed into an empty 500 g glass bottle. What should the minimum temperature of the bottle be in order for all the ice to melt? (5 points) (Calass = 840 J/kg.°C, Lv.ice = 333 kJ/kg)

Answer 1

To be given in question

$(C_{v})_{ice}=2.108\times10^3jule/kg -K$

$(C_{v})_{w}=4.187\times10^3Jule/kg -K$

$m_{1}=$ 2.20 gm

$T_{1}=0°C$

$m_{2}=500gm$

$[C_{v}]_{glass}=840J/kg°C$

$[{L_{v}}]_{ice}=333kJ/kg$

Point= 5

To be asked in question

T₂=?

We know that

$(mc\Delta T)_{ice} +(mL)_{ice}=(mc\Delta T)_{glass}$

$2.20\times2.108\times10^3(0-T) +2.20\times333\times10^3=500\times4.187\times10^3(0-T)$

$We can solve$

$2.20\times2.108(0-T) +2.20\times333=500\times4.187(0-T)$

T=3.505°C