Question

2. Imagine now that one tried to repeat the same measurement, but between the time that the warm water was measured and the time it was mixed with the cooler water, the warmer water has lost 3060 cal to the environment. Calculate the final equilibrium temperature assuming no additional heat loss after mixing occurs.

Answer 1

Answer:

The final temperature is 64.28°C.

Explanation:

Imagine now that one tried to repeat the same measurement, but between the time that the 400 cm³ of warm water was measured at 88°C and the time it was mixed with the 130 cm³ of cooler water at 15°C, the warmer water had lost 3060 cal to the environment. Calculate the final equilibrium temperature assuming no heat loss after mixing occurs.

Let the temperature of hot water before mixing is t

We need to calculate the temperature

Using formula of energy

$Q=mc\Delta t$

$Q=V\times\rho\times c\times(T_{2}-T_{1})$

Put the value into the formula

$3060=400\times1\times1\times(88-t)$

$3060=400\times88-400t$

$-t=\dfrac{3060-400\times88}{400}$

$t=80.35^{\circ}$

We need to calculate the final temperature

Using formula of temperature

$T_{f}=\dfrac{V_{w}T_{w}+V_{c}T_{c}}{V_{w}+V_{c}}$

Put the value into the formula

$T_{f}=\dfrac{400\times80.3+130\times15}{400+130}$

$T_{f}=64.28^{\circ}C$

Hence, The final temperature is 64.28°C.