Answer:
(a) the mass of the water is 3704 g
(b) the mass of the water is 199, 285.7 g
Explanation:
Given;
Quantity of heat, H= 8.37 x 10⁶ J
Part (a) mass of water (as sweat) need to evaporate to cool that person off
Latent heat of vaporization of water, Lvap. = 2.26 x 10⁶ J/kg
H = m x Lvap.
mass in gram ⇒ 3.704 kg x 1000g = 3704 g
Part (b) quantity of water raised from 25.0 °C to 35.0 °C by 8.37 x 10⁶ J
specific heat capacity of water, C, 4200 J/kg.°C
H = mcΔθ
where;
Δθ is the change in temperature = 35 - 25 = 10°C
mass in gram ⇒ 199.2857 kg x 1000 g = 199285.7 g
Answer:
60 grams of ice will require 30.26 calories to raise the temperature 1°C.
Explanation:
The amount of heat (Q) to raise the temperature of 60.0 g of ice by 1°C can be calculated from:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat released or absorbed by the system.
m is the mass of the ice (m = 60.0 g).
c is the specific heat capacity of ice (c = 2.108 J/g.°C).
ΔT is the temperature difference (ΔT = 1.0 °C).
∴ Q = m.c.ΔT = (60.0 g)(2.108 J/g.°C)(1.0 °C) = 126.48 J.
<em>It is known that 1.0 cal = 4.18 J.</em>
<em>∴ Q = (126.48 J)(1.0 cal / 4.18 J) = 30.26 cal.</em>
There the picture is the answer from the internet
Answer:
0.486 L
Explanation:
Step 1: Write the balanced reaction
2 KCIO₃(s) ⇒ 2 KCI (s) + 3 O₂(g)
Step 2: Calculate the moles corresponding to 1.52 g of KCIO₃
The molar mass of KCIO₃ is 122.55 g/mol.
1.52 g × 1 mol/122.55 g = 0.0124 mol
Step 3: Calculate the moles of O₂ produced from 0.0124 moles of KCIO₃
The molar ratio of KCIO₃ to O₂ is 2:3. The moles of O₂ produced are 3/2 × 0.0124 mol = 0.0186 mol
Step 4: Calculate the volume corresponding to 0.0186 moles of O₂
0.0186 moles of O₂ are at 37 °C (310 K) and 0.974 atm. We can calculate the volume of oxygen using the ideal gas equation.
P × V = n × R × T
V = n × R × T/P
V = 0.0186 mol × (0.0821 atm.L/mol.K) × 310 K/0.974 atm = 0.486 L
