# Problem: A 30.0-g sample of water at 280. K is mixed with 50.0 g water at 330. K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings.

###### FREE Expert Solution

We’re being asked to determine the final temperature of the given mixture.

We will use the heat released by the sample of steam to calculate the final temperature of the mixture.

Recall that heat (q) can be calculated using the following equation:

$\overline{){\mathbf{q}}{\mathbf{=}}{\mathbf{mc}}{\mathbf{∆}}{\mathbf{T}}}$

q = heat, J

+qabsorbs heat
–qloses heat

m = mass (g)
c = specific heat capacity = J/(g·°C)
ΔT = Tf – Ti = (°C)

Given:

For q1 (@ 330. K)

mass= 50.0 g
c = 4.18 J/(gK)
Ti = 330 K

For q2 (@ 280. K)

mass= 30 g
c = 4.18 J/(gK)
Ti = 280 K

heat lost by water at 330 K = heat absorbed by water at 280 K

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###### Problem Details

A 30.0-g sample of water at 280. K is mixed with 50.0 g water at 330. K. Calculate the final temperature of the mixture assuming no heat loss to the surroundings.