Answer:
T(final) = 40.4 °C
Explanation:
Given data:
Density of water = 1.00 g/mL
Volume of water at 25°C = 390.0 mL
Volume of water at 95°C = 110.0 mL
Final temperature of mixture = ?
Solution:
Heat absorbed by warm water = heat lost by hot water
q lost = -q lost ....... (1)
Formula:
q = m.c.ΔT
ΔT(warm) = T(final) - 25°C
ΔT(hot) = T(final) - 95°C
Mass of water:
d = m/v
1 g/mL = m/ 390 mL
390 g = m
For hot water:
d = m/v
1 g/mL = m/ 110.0 mL
110.0 g = m
Now we will put the values in equation 1.
390 g. c. (T(final) - 25°C) = - 110.0 g .c. (T(final) - 95°C)
390 T(final) - 9750 °C = -110 T(final) + 10450 °C
390 T(final) +110 T(final) = 10450 °C +9750 °C
500 T(final) = 20200
T(final) = 20200/500
T(final) = 40.4 °C
