Question

A jar of tea is placed in sunlight until it

reaches an equilibrium temperature of 33.3
◦C .
In an attempt to cool the liquid, which has a
mass of 187 g , 133 g of ice at 0.0
◦C is added.
At the time at which the temperature of the
tea is 31.8
◦C , find the mass of the remaining
ice in the jar. The specific heat of water
is 4186 J/kg ·
◦ C . Assume the specific heat
capacity of the tea to be that of pure liquid
water.
Answer in units of g.
(2 significant digits)

Answer 1

To solve this problem, we need to use the following formula:

Q = m_tea * c_tea * (T_f - T_i) + m_ice * L_f + m_ice * c_ice * (T_f - 0)

where Q is the amount of heat transferred, m_tea is the mass of the tea, c_tea is the specific heat capacity of the tea, T_i is the initial temperature of the tea, T_f is the final temperature of the tea and ice mixture, m_ice is the mass of the ice, L_f is the latent heat of fusion of ice (334 J/g), and c_ice is the specific heat capacity of ice (2.108 J/g·°C).

First, we need to calculate the initial temperature of the tea. Since it has reached an equilibrium temperature of 33.3°C in sunlight, we can assume that its initial temperature was also 33.3°C.

So, the equation becomes:

Q = (187 g) * (4186 J/kg·°C) * (31.8°C - 33.3°C) + (133 g) * (334 J/g) + (m_ice) * (2.108 J/g·°C) * (31.8°C - 0°C)

Simplifying this equation, we get:

Q = -121732.8 J + 44422 J + 67.032 m_ice

Setting Q to zero, since we want to find the mass of the remaining ice when the temperature is 31.8°C, we get:

67.032 m_ice = 121732.8 J - 44422 J

m_ice = 114.9 g

Therefore, the mass of the remaining ice in the jar when the temperature is 31.8°C is 114.9 g (to 2 significant digits).