A jar of tea is placed in sunlight until it reaches an equilibrium temperature of 31.8 ◦C . In an attempt to cool the liquid, which has a mass of 161 g , 131 g of ice at 0.0 ◦C is added. At the time at which the temperature of the tea is 28.8 ◦C , find the mass of the remaining

Explanation:

mass of liquid m₁ = 161 g

temperature t₁ = 31.8

final temperature t₂ = 28.8

Let m g of ice melted to cool the liquid

heat gained = mass x latent heat of fusion + mass x loss of temp x s heat of water

= m x 80 + m x 1 x ( 31.8 - 28.8 ) ( latent heat of ice = 80 cals/g )

= 83 m

heat lost  = 161 x 1 x ( 31.8 - 28.8 ) ( specific heat of water = 1 cal / g / k )

= 161 x 3

heat lost = heat gained

83 m = 161 x 3

m = 5.82 g

mass of remaining ice = 131 - 5.82

= 125.18 g

battery with hi vtge high val resist in series