Answer:
The hot tea should transfer <em>25.63 kJ</em> the surroundings to cool the tea.
Explanation:
- To solve this problem, we can use the relation:
<em>Q = m.c.ΔT,</em>
where, Q is the amount of heat has to be transferred from the tea to the surroundings to cool the tea (Q = ??? J).
m is the mass of the hot tea (m = dV = (1.0 g/mL)(250 mL) = 250 g), suppose the density of water is the density of tea.
c is the specific heat of the hot tea (c = 4.10 J/°C.g).
ΔT is the difference between the initial and final temperature (ΔT = final T - initial T = 350 K - 375 K = -25°C).
<em>∴ Q = m.c.ΔT</em> = (250 g)(4.10 J/°C.g)(-25°C)) = <em>- 25630 J = - 25.63 kJ.</em>
<em>So, the hot tea should transfer 25.63 kJ the surroundings to cool the tea.</em>
Percent error=(experimentally determined boiling point-actual boiling point)
=(124.1 - 125.7) = 1.6 °C
Quantity Number of significant figures
0.070020 meter 5
10,800 meters 5
0.010 square meter 2
5.00 cubic meters 1
507 thumbtacks 3
(5.3 x 10⁴) + (1.3 x 10⁴) =6.6 x 10⁴
(7.2 x 10⁻⁴) ÷ (1.8 x 10³)=4 x 10⁻⁷
10⁴ x 10⁻³ x 10⁶=10⁷
(9.12 x 10⁻¹) - (4.7 x 10⁻²)=(9.12 x 10⁻²) - (4.7 x 10⁻²)
=(91.2-4.7) x 10⁻²
=85.5 x 10⁻²
Potential cause it holding the force?
