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>
Answer:
8 grams.
Explanation:
When solving this type of problem, you need to find how many grams 4.0% of 200g will be equivalent to. You can do this by setting up an equation as such:
200 grams ×0.04 = 8 grams. (0.04 is equivalent to 4%)
