# 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

90% (389 ratings) ###### 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.

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Thermal Equilibrium concept. You can view video lessons to learn Thermal Equilibrium. Or if you need more Thermal Equilibrium practice, you can also practice Thermal Equilibrium practice problems.

What textbook is this problem found in?

Our data indicates that this problem or a close variation was asked in Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition. You can also practice Chemistry: An Atoms First Approach - Zumdahl Atoms 1st 2nd Edition practice problems.