The final temperature of the mixture in the coffee cup calorimeter is; 19.467 °C
According to the law of energy conservation:
As such; the heat transfer in the liquid water is equal to heat gained by the ice
Heat transfer by liquid water is therefore;
- DH = 6.02 kJ/mol) = 150 × 4.18 × (T1 - T2)
- 6020 J/mol = 627 × (20 - T2)
However, since 18g of water makes one mole
- 6020 J/mol = 6020/18 = 334.44 J/g.
T2 = 19.467°C
Read more on heat:
brainly.com/question/21406849
1.00 degree increase<span> on the </span>Celsius scale<span> is </span>equivalent<span> to a </span>1.80 degree increase<span> on the</span>Fahrenheit scale<span>. If a </span>temperature increases<span> by </span>48.0°C<span>, </span>
Answer: -
1595 g
Explanation: -
Heat of vaporization = 2.4 Kj/ g
Mass of water to be vaporized = 50 g
Heat released = Mass of water to be vaporized x Heat of vaporization
= 50 g x 2.4 KJ /g
= 120 KJ
= 120000 J
Initial temperature= 33+273= 306 K
Final temperature =15+273=288 K
Change in temperature required = T = 306 - 288 = 18 K
specific heat of water is 4.18 J / g K
Mass of water that can be cooled = Total heat / (specific heat of water x Change in temperature)
= 120000 J / ( 4.18 J / g K x 18 K)
= 1595 g
Since the temperature
is a constant, we can use Boyle's law to solve this.<span>
<span>Boyle' law says "at a constant temperature, the
pressure of a fixed amount of an ideal gas is inversely proportional to its
volume.
P α 1/V
</span>⇒
PV = k (constant)<span>
Where, P is the pressure of the gas and V is the
volume.
<span>Here, we assume that the </span>gas in the balloon is an ideal gas.
We can use Boyle's law for these two situations as,
P</span>₁V₁ = P₂V₂<span>
P₁ = 100.0 kPa = 1 x 10⁵ Pa
V₁ =
3.3 L
P₂ =
90.0 x 10³ Pa
V₂ =?
By substitution,
1 x 10⁵ Pa x 3.3 L = 90 x 10³ Pa x V₂</span><span>
V</span>₂ = 3.7 L<span>
</span><span>Hence, the volume of gas when pressure is 90.0 kPa
is 3.7 L.</span></span>
