Question

If 700 g of water at 90 °C loses 27 kJ of heat, what is its final temperature?​

Answer 1

Answer:

If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C

Explanation:

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

In this way, between heat and temperature there is a direct proportional relationship (Two magnitudes are directly proportional when there is a constant so that when one of the magnitudes increases, the other also increases; and the same happens when either of the two decreases .). The constant of proportionality depends on the substance that constitutes the body and its mass, and is the product of the specific heat and the mass of the body. So, the equation that allows to calculate heat exchanges is:

Q = c * m * ΔT

Where Q is the heat exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature, ΔT= Tfinal - Tinitial

In this case:

• Q= 27 kJ= 27,000 J (being 1 kJ=1,000 J)
• $c=4.186 \frac{J}{g* C}$
• m=700 g
• ΔT= Tfinal - Tinitial= Tfinal - 90 °C

Replacing:

$27,000 J=4.186 \frac{J}{g* C}*400 g* (Tfinal - 90C)$

Solving:

$27,000 J=1,674.4 \frac{J}{C}* (Tfinal - 90C)$

$\frac{27,000 J}{1,674.4 \frac{J}{C}} =(Tfinal - 90C)$

16.125 °C= Tfinal - 90 °C

Tfinal= 16.125 °C + 90 °C

Tfinal= 106.125 °C

If 700 g of water at 90 °C loses 27 kJ of heat, its final temperature is 106.125 °C