If you combine 390.0 ml of water at 25.00 °c and 110.0 ml of water at 95.00 °c, what is the final temperature of the mixture? use 1.00 g/ml as the density of water.

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

possibly, yes.

explanation:

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