Question

A 100.0 g sample of water at 27.0°c is poured into a 70.3 g sample of water at 89.0°c. what will be the final temperature of the water?

Answer 1

100g*4.186(specific heat of water)(tf-27)=-70.3g(negative)*(4.186)*(tf-89)
that should do

Answer 2

Answer:

The final temperature of water = 52.6 C

Explanation:

The heat (q) lost or gained by a substance of mass m corresponding to a temperature change from T1 to T2 degrees is given as:

$q = m*c*(T2-T1)$-----(1)

where c = specific heat of the substance

For water, c = 4.18 J/gC

In the given situation:

Heat lost by 70.3 g of water = Heat gained by 100.0 g of water

Based on equation (1) and considering that heat lost is negative:

$-70.3*c*(T2-89.0)=100.0*c*(T2-27.0)$

Solving the above equation gives:

T2 = 52.6 C