Answer: 1560632 joules
Explanation:
The change in thermal energy (Q) required to heat ice depends on its Mass (M), specific heat capacity (C) and change in temperature (Φ)
Thus, Q = MCΦ
Given that:
Q = ?
Mass of frozen water (ice) = 1kg
C = 4184 J/(kg K)
Φ = (Final temperature - Initial temperature)
= 100°C - 0°C = 100°C
Convert 100°C to Kelvin
(100°C + 273) = 373K
Then, Q = MCΦ
Q = 1kg x 4184 J/(kg K) x 373K
Q = 1560632 joules
Thus, the change in thermal energy is 1560632 joules
Answer:
248 mL
Explanation:
According to the law of conservation of energy, the sum of the heat absorbed by water (Qw) and the heat released by the coffee (Qc) is zero.
Qw + Qc = 0
Qw = -Qc [1]
We can calculate each heat using the following expression.
Q = c × m × ΔT
where,
- ΔT: change in the temperature
163 mL of coffee with a density of 0.997 g/mL have a mass of:
163 mL × 0.997 g/mL = 163 g
From [1]
Qw = -Qc
cw × mw × ΔTw = -cc × mc × ΔTc
mw × ΔTw = -mc × ΔTc
mw × (54.0°C-25.0°C) = -163 g × (54.0°C-97.9°C)
mw × 29.0°C = 163 g × 43.9°C
mw = 247 g
The volume corresponding to 247 g of water is:
247 g × (1 mL/0.997 g) = 248 mL
Answer:
i think so
Ba(OH)2 + H2SO4 ------> BaSO4 + 2H2O
1) Moles of Ba(OH)2 = moles of H2SO4 = 0.025L x 2)0.02M = 5.0 x 10^-4M
Concn of Ba(OH)2 in g/L = 5.0 x 10^-4M x 171.33g/mol = 0.086g/mol
