Question

unit 3 - quiz

1. on a counter is a glass of water with ice cubes floating in it. you measure the temperature and find it to be 0°c. would the temperature of the water change if you were to add more ice cubes to the glass? explain your answer.

1. suppose that 1500 kj of energy were transferred to water at 20°c. what mass of water could be brought to the boiling point? heat capacity (c) for liquid water is 4.18 j/g°c

1. an ice cube (25 g) is at -8.0°c. how much energy is required to completely melt it? heat capacity (c) for solid water is 2.10 j/g°c. h_f = 334 j/g

energy constants (h₂o)
334 j/g heat of fusion (melting or freezing) h_f
2260 j/g heat of vaporization (evaporating or condensing) h_v
2.1 j/g°c heat capacity (c) of solid water
4.18 j/g°c heat capacity (c) of liquid water

q = m·c·δt
q = m·h_v or m·h_f

Explanation:

Question 2

Step1: Identify known values

We know the energy transferred \( Q = 1500\space kJ = 1500\times1000\space J = 1500000\space J \), initial temperature \( T_i = 20^\circ C \), boiling point of water \( T_f = 100^\circ C \), and specific heat capacity of water \( c = 4.18\space J/g^\circ C \).

Step2: Use the heat formula \( Q = m\cdot c\cdot\Delta T \)

First, calculate the temperature change \( \Delta T = T_f - T_i = 100 - 20 = 80^\circ C \). Then, rearrange the formula to solve for mass \( m \): \( m=\frac{Q}{c\cdot\Delta T} \).

Step3: Substitute the values

Substitute \( Q = 1500000\space J \), \( c = 4.18\space J/g^\circ C \), and \( \Delta T = 80^\circ C \) into the formula: \( m=\frac{1500000}{4.18\times80} \).

Step4: Calculate the mass

First, calculate the denominator: \( 4.18\times80 = 334.4 \). Then, \( m=\frac{1500000}{334.4}\approx4485.65\space g \) or approximately \( 4.49\space kg \).

Step1: Energy to heat ice to \( 0^\circ C \)

The ice is at \( - 8.0^\circ C \) and needs to be heated to \( 0^\circ C \). Use the formula \( Q_1=m\cdot c\cdot\Delta T \), where \( m = 25\space g \), \( c = 2.10\space J/g^\circ C \), and \( \Delta T=0 - (-8)=8^\circ C \). So, \( Q_1 = 25\times2.10\times8 \).

Step2: Calculate \( Q_1 \)

\( Q_1=25\times2.10\times8 = 25\times16.8 = 420\space J \).

Step3: Energy to melt the ice

Use the formula \( Q_2=m\cdot H_f \), where \( m = 25\space g \) and \( H_f = 334\space J/g \). So, \( Q_2=25\times334 \).

Step4: Calculate \( Q_2 \)

\( Q_2 = 25\times334=8350\space J \).

Step5: Total energy required

Total energy \( Q = Q_1 + Q_2=420 + 8350 \).

Answer:

The mass of water is approximately \(\boldsymbol{4486\space g}\) (or \(\boldsymbol{4.49\space kg}\))

Question 3