Question

One liter of water at 56◦C is used to make iced

tea.
How much ice at 0 ◦C must be added to
lower the temperature of the tea to 20 ◦C?
The specific heat of water is 1 cal/g ·
◦ C and
latent heat of ice is 79.7 cal/g.
Answer in units of g.

Answer 1

To lower the temperature of the tea from 56◦C to 20◦C, we need to calculate the amount of heat that needs to be removed. By setting up an equation and solving for the mass of ice, we find that 1.8 g of ice at 0 ◦C must be added.

Step-by-step explanation:

To lower the temperature of the tea from 56◦C to 20◦C, we need to calculate the amount of heat that needs to be removed. First, we need to calculate the heat absorbed by the ice to melt it. The specific heat of water is 1 cal/g · ◦ C, so the heat absorbed by the ice can be calculated using the formula Q = m × s × ΔT, where Q is the heat absorbed, m is the mass of ice, s is the specific heat of water, and ΔT is the change in temperature. In this case, ΔT is 0 - (-20) = 20 ◦C, and the specific heat of water is 1 cal/g · ◦ C. Therefore, the heat absorbed by the ice is given by Q = m × 1 cal/g · ◦ C × 20 ◦C = 20m cal.

Next, we need to calculate the heat removed from the tea. The specific heat of water is 1 cal/g · ◦ C, so the heat removed from the tea can be calculated using the same formula Q = m × s × ΔT, where Q is the heat removed, m is the mass of water, s is the specific heat of water, and ΔT is the change in temperature. In this case, ΔT is 56 - 20 = 36 ◦C, and the specific heat of water is 1 cal/g · ◦ C. Therefore, the heat removed from the tea is given by Q = m × 1 cal/g · ◦ C × 36 ◦C = 36m cal.

Since the heat absorbed by the ice is equal to the heat removed from the tea, we can set up the equation 20m cal = 36m cal and solve for m. Dividing both sides of the equation by 20 cal, we get m = 36/20 = 1.8 g. Therefore, 1.8 g of ice at 0 ◦C must be added to lower the temperature of the tea to 20 ◦C.

Answer 2

m= 361.0832 g

Step-by-step explanation:

This answer is identitical to the answer before me except you add together the latent heat of ice and the temperature change of ice instead of multiplyng them.

Mass of water(mw)=1liter=1000g

Final temperature=20°C

Temperature of ice=0°C

Temperature of water=56°C

Change in temperature of water=56-20=36

change in temperature of ice=20-0=20

Specific heat of water=1cal/g°C

Latent heat of ice=79.7cal/g

1000x1x36=mx(79.7+20) <- only difference from the answer above

36000=99.7xm

Divide both sides by 99.7

22.6=m of ice

m of ice=361.0832g