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

Answer: The mass of the remaining ice in the jar is 1.3e+2 g.

Explanation: Let’s denote the mass of the remaining ice as m_ice. The heat gained by the ice is equal to the heat lost by the tea. The heat gained by the ice is given by m_ice * L_f, where L_f is the latent heat of fusion of water (334000 J/kg). The heat lost by the tea is given by m_tea * c_w * (T_initial - T_final), where m_tea is the mass of tea (0.187 kg), c_w is the specific heat capacity of water (4186 J/kg·°C), T_initial is the initial temperature of the tea (33.3°C), and T_final is the final temperature of the tea (31.8°C).

Equating the heat gained by the ice to the heat lost by the tea, we get:

m_ice * L_f = m_tea * c_w * (T_initial - T_final)

Substituting in the values, we get:

m_ice * 334000 = 0.187 * 4186 * (33.3 - 31.8)

Solving for m_ice, we get:

m_ice = 0.187 * 4186 * (33.3 - 31.8) / 334000

m_ice ≈ 0.130 kg

Converting to grams and rounding to two significant figures, we get:

m_ice ≈ 130 g

Hope this helps, and have a great day! =)