Ice at -20° C is mixed with 50g of water at 40° C . When the temperature of the mixture reaches 0° C , it is found that 20g of ice is still unmelted. choose the closest value of amount of ice added to the water from given option (Specific heat of water =4.2J/g/° C Specific heat of Ice=2.1J/g/° C Heat of fusion of water at 0° C=334J/g

Question

Ice at -20° C is mixed with 50g of water at 40° C . When the temperature of the mixture reaches 0° C , it is found that 20g of ice is still unmelted. choose the closest value of amount of ice added to the water from given option (Specific heat of water =4.2J/g/° C Specific heat of Ice=2.1J/g/° C Heat of fusion of water at 0° C=334J/g (A) 100g (B) 40g (C) 60g (D) 50g

Tardigrade Admin · Accepted Answer

Heat lost by water Δ Q=msΔ T=50× 1× 40=2000 text cal Let mass of ice =m Heat gain by ice Δ Q=m× (1/2)× 20+(.m-20.)× 80 =10m+80m-1600=90m-1600 Heat gain = Heat lost 90m-1600=2000 m=40gm

$100 g$

$40 g$

$60 g$

$50 g$

Heat lost by water $Δ Q = m s Δ T = 50 \times 1 \times 40 = 2000 cal$
Let mass of ice $= m$
Heat gain by ice $Δ Q = m \times \frac{1}{2} \times 20 + (m - 20) \times 80$
$= 10 m + 80 m - 1600 = 90 m - 1600$
Heat gain $=$ Heat lost
$90 m - 1600 = 2000$
$m = 40 g m$

100g

40g

60g

50g

Heat lost by water ΔQ=msΔT=50×1×40=2000 cal Let mass of ice =m Heat gain by ice ΔQ=m×21​×20+(m−20)×80 =10m+80m−1600=90m−1600 Heat gain = Heat lost 90m−1600=2000 m=40gm

$100 g$

$40 g$

$60 g$

$50 g$

Heat lost by water $Δ Q = m s Δ T = 50 \times 1 \times 40 = 2000 cal$
Let mass of ice $= m$
Heat gain by ice $Δ Q = m \times \frac{1}{2} \times 20 + (m - 20) \times 80$
$= 10 m + 80 m - 1600 = 90 m - 1600$
Heat gain $=$ Heat lost
$90 m - 1600 = 2000$
$m = 40 g m$