Question

A 20g sample of iron at a temperature of 120oC is placed into a container of water. There are 300 milliliters of water in the container at a temperature of 30oC. Ciron=0.444Jg∘C Cwater=4.184Jg∘C rhowater=1gmL 10. What is the final temperature of the water?

Answer 1

Answer:

30.63 °C will be the final temperature of the water.

Explanation:

Heat lost by iron will be equal to heat gained by the water

$-Q_1=Q_2$

Mass of iron = $m_1=20 g$

Specific heat capacity of iron = $c_1=0.444 J/g^oC$

Initial temperature of the iron = $T_1=120^oC$

Final temperature = $T_2$=T

$Q_1=m_1c_1\times (T-T_1)$

Volume of water = 300 ml

Density of water = 1 g/mL

Mass of water= $m_2=300 mL\times 1 g/mL = 300 g$

Specific heat capacity of water= $c_2=4.184 J/g^oC$

Initial temperature of the water = $T_3=30^oC$

Final temperature of water = $T_2$=T

$Q_2=m_2c_2\times (T-T_3)$

$-Q_1=Q_2$

$-(m_1c_1\times (T-T_1))=m_2c_2\times (T-T_3)$

On substituting all values:

we get, T =  30.63°C

30.63 °C will be the final temperature of the water.