Answer:
The hot tea should transfer 25.63 kJ the surroundings to cool the tea.
Step-by-step explanation:
- To solve this problem, we can use the relation:
Q = m.c.ΔT,
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).
∴ Q = m.c.ΔT = (250 g)(4.10 J/°C.g)(-25°C)) = - 25630 J = - 25.63 kJ.
So, the hot tea should transfer 25.63 kJ the surroundings to cool the tea.