### Calculate the final temperature when a 136 mL of H_{2}O at 25.0 C
is mixed with 75.0 mL of H_{2}O at 82.0 C.

### Solution:

### The important thing to remember here is when the two samples of water are
mixed the sample at the higher temperature will transfer heat to the sample
at the lower temperature. We can write this mathematically as,

### q_{hot water} = -q_{cold water}

### The amount of heat transferred is determined by the equation;

### q= S.H. * mass * T

### So substituting this relationship into the first equation we get,

### S.H._{hot water} * mass_{hot water} * T_{hot
water = - S.H.cold water * masscold water * Tcold
water}

### Now we simply substitute the information we know, (remember the
density of water is 1.00 g mL^{-1}

### 1.84 J·g^{-1}C^{-1}
* 136 g_{} * T_{hot water}
= - 1.84 J·g^{-1}C^{-1} * 75.0
g_{} * T_{cold water}

### This can be reduced to,

### 1.81 (T_{hot
water}) = - T_{cold water}

### Remember that,

### T
= T_{final} - T_{initial}

### Therefore,

### 1.81 (T_{final} - T_{initial})_{hot} = - (T_{final} - T_{initial})_{cold}

### Substituting;

### 1.81 (T_{final} - 25 C_{})_{hot} =
- (T_{final} - 82.0 C)_{cold}

### 1.81T_{final} - 45.25 = - T_{final} + 82.0
C

### 2.81T_{final} = 127.2

### T_{final} = 45.3 C

###

###

###

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