A30-0 g sample of water at 280 k is mixed with 50.0 g of water at 330 k. how would you calculate the final temperature of the mixture assuming no heat is lost to the surroundings?

311.25k

Explanation:

The question assumes heat is not lost to the surroundings, therefore

heat emitted from hotter sample ( )= heat absorbed by the less hotter  sample( )

The relationship between heat (q), mass (m) and temperature (t) is

where c is specific heat capacity, temperature change.

=

equating both heat emitted and absorb

where the values with subset 1 are the values of the hotter sample of water and the values with subset 2 are the values of the less hot sample of water.

C will cancel out since both are water and they have the same specific heat capacity.

so we have

where m1 = 50g, t 1initial = 330, m2 = 30g, t2 initial = 280,t final (final temperature of the mixture) = ?

-50 * ( - 330) = 30 *  ( - 280)

-50 + 16500 = 30 - 8400

80 = 16500+8400

80 = 24900

= 24900/80 = 311.25k

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