Question

for each of the following, select only the correct answer corresponding to the question on the mcq card provided.

1. sandile conducts a calorimetry experiment and fills a glass beaker of unknown mass with 74 ml of water. he puts the beaker with water on a hot plate where it absorbs 2000 calories of heat and the temperature of the system increases by 20°c. given that the specific heat capacity of water is 1.0 cal/g·°c and that of glass is 0.18 cal/g·°c, the mass of the beaker is closest to

a. 270 000 g
b. 140 g
c. 560 g
d. 540 g

1. a solid substance will turn into liquid when

a. its temperature increases.
b. its temperature decreases.
c. heat is removed from it.
d. heat is added to it.

1. how much heat must be removed from 456 g of water at 25.0°c to change it into ice at - 10.0°c? the specific heat of ice is 2100 j/kg·°c, the latent heat of fusion of water is 33.5 × 104 j/kg, and the specific heat of water is 4186 j/kg·°c.

a. 105 kj
b. 153 kj
c. 57.3 kj
d. 210 kj

Explanation:

Question 1

Step1: Calculate the mass of water

Density of water \(
ho = 1\ g/mL\). Given \(V = 74\ mL\), using \(m=
ho V\), we have \(m_{water}=74\ g\)

Step2: Use the heat - transfer formula \(Q = Q_{water}+Q_{beaker}\)

The heat - transfer formula is \(Q = m_{water}c_{water}\Delta T+m_{beaker}c_{beaker}\Delta T\). We know \(Q = 2000\ cal\), \(c_{water}=1.0\ cal/g^{\circ}C\), \(c_{beaker}=0.18\ cal/g^{\circ}C\), \(\Delta T = 20^{\circ}C\) and \(m_{water}=74\ g\)
Substitute the values into the formula: \(2000=74\times1.0\times20 + m_{beaker}\times0.18\times20\)

Step3: Solve for \(m_{beaker}\)

First, calculate \(74\times1.0\times20=1480\). Then the equation becomes \(2000 = 1480+3.6m_{beaker}\).
Subtract 1480 from both sides: \(3.6m_{beaker}=2000 - 1480=520\).
Then \(m_{beaker}=\frac{520}{3.6}\approx144.4\ g\approx140\ g\)

A solid substance turns into a liquid through the process of melting. Melting is an endothermic process, which means heat is added to the substance. When heat is added, the kinetic energy of the particles in the solid increases. At the melting point, the added heat overcomes the intermolecular forces holding the particles in a fixed position (in the solid state), and the substance changes to a liquid state.

Step1: Calculate the heat removed to cool water from \(25^{\circ}C\) to \(0^{\circ}C\)

\(m = 456\ g=0.456\ kg\), \(c_{water}=4186\ J/kg^{\circ}C\), \(\Delta T_1=25 - 0=25^{\circ}C\)
Using \(Q_1 = mc\Delta T\), we get \(Q_1=0.456\times4186\times25=47794\ J\)

Step2: Calculate the heat removed to freeze water at \(0^{\circ}C\)

\(L_f = 33.5\times 10^{4}\ J/kg\). Using \(Q_2=mL_f\), we have \(Q_2=0.456\times33.5\times 10^{4}=152960\ J\)

Step3: Calculate the heat removed to cool ice from \(0^{\circ}C\) to \(- 10^{\circ}C\)

\(c_{ice}=2100\ J/kg^{\circ}C\), \(\Delta T_2=0-(-10) = 10^{\circ}C\)
Using \(Q_3=mc\Delta T\), we get \(Q_3=0.456\times2100\times10 = 9576\ J\)

Step4: Calculate the total heat removed \(Q = Q_1+Q_2+Q_3\)

\(Q=47794 + 152960+9576=210330\ J\approx210\ kJ\)

Answer:

B. \(140\ g\)

Question 2