Question

1. which statement is true?

a. room temperature is approximately 23 °k
b. \absolutely zero\ refers to 0 °f
c. a change of 1 °k is equivalent to a change of 1 °c
d. there can be a negative temperature in °c and in °k
e. none of the above

1. using the heat curve to the right, what is true about the substance?

a. the substance has a melting point of approximately 200 °c
b. the substance has a boiling point of approximately 200 °c
c. the substance has a melting point of approximately 800 °c
d. the substance has a boiling point of approximately 800 °c
e. the substances latent heat of vaporization is greater than its latent heat of fusion.

1. what is the change in heat energy when 8.5 kg of water is cooled from 85°c to 35°c? use

( c _ { \text { water } } = 4186 mathrm { j } / mathrm { kg } )
a. ( 1.8 \times 10 ^ { 6 } mathrm { j } ) b. ( - 1.8 \times 10 ^ { 6 } mathrm { j } ) c. ( 3.0 \times 10 ^ { 6 } mathrm { j } ) d. ( 1.2 \times 10 ^ { 6 } mathrm { j } ) e. ( 0 mathrm { j } )
short answer problems: show your work where appropriate. 0.5 communication mark will be deducted for each
missing incorrect unit, direction or significant figure, up to 3 marks. 3 marks - c
error tracker, do not write on these lines: #13 _ #14 _ #15 _ #16 _

1. a 3.2 kg sample of iron at a temperature of 175°c is placed in 12.0 l of 20.0 °c water. what will be the

temperature of the system if allowed to sit? ( c _ { \text { water } } = 4186 mathrm { j } / mathrm { kg } ^ { circ } mathrm { c } ; c _ { \text { iron } } = 450 mathrm { j } / mathrm { kg } ^ { circ } mathrm { c } ) 4 marks - t

1. danielle cools a 2.0 kg metal object to a temperature of -25.0 °c. she places the metal in 3.0 l of pure water

initially at a temperature of 40.0 °c. the final temperature of the mixture is 36.0 °c. what is the specific heat
capacity of the metal? 4 marks - a

Explanation:

Step1: Calculate the mass of water

The density of water is \(1000\ kg/m^3 = 1\ kg/L\). So, for \(V = 12.0\ L\) of water, \(m_{water}=
ho V=12.0\ kg\)

Step2: Set up the heat - transfer equation

According to the principle of conservation of energy \(Q_{lost}=Q_{gained}\). The heat - transfer formula is \(Q = mc\Delta T\).
For iron: \(Q_{iron}=m_{iron}c_{iron}(T_{iron}-T)\)
For water: \(Q_{water}=m_{water}c_{water}(T - T_{water})\)
Since \(Q_{iron}=Q_{water}\), we have \(m_{iron}c_{iron}(T_{iron}-T)=m_{water}c_{water}(T - T_{water})\)
Substitute \(m_{iron} = 3.2\ kg\), \(c_{iron}=450\ J/kg^{\circ}C\), \(T_{iron}=175^{\circ}C\), \(m_{water}=12.0\ kg\), \(c_{water}=4186\ J/kg^{\circ}C\), \(T_{water}=20.0^{\circ}C\) into the equation:

$$3.2\times450\times(175 - T)=12.0\times4186\times(T - 20.0)$$

Step3: Expand the equation

$$3.2\times450\times175-3.2\times450T=12.0\times4186T-12.0\times4186\times20.0$$

$$3.2\times450\times175 + 12.0\times4186\times20.0=(12.0\times4186 + 3.2\times450)T$$

Calculate \(3.2\times450\times175=3.2\times78750 = 252000\)
Calculate \(12.0\times4186\times20.0=12.0\times83720 = 1004640\)
Calculate \(12.0\times4186+3.2\times450=50232 + 1440=51672\)
Then \(T=\frac{252000 + 1004640}{51672}\)

Step4: Solve for \(T\)

$$T=\frac{1256640}{51672}\approx24.3^{\circ}C$$

Answer:

The final temperature of the system is approximately \(24.3^{\circ}C\)