Question

Question:A 5.00-g sample of aluminium pellets (specific heat capacity =0.89 J ⁰C^-1 g^-1) and a 10.00-g sample of iron pellets (specific heat capacity = 0.45 J ⁰C^-1 g^-1) are heated to 100.0⁰C. The mixture of hot iron and aluminium is then dropped into 97.3 g of water at 22.0⁰C. Calculate the final temperature of the metal and water mixture, assuming no heat loss to the surroundings.

Short answer

Thefinal temperature of the metal and water mixturewill be 23.7⁰C

Step-by-step solution

Given data

From the given information we can write

Heat gain by water=Heat loss by Al+ Heat loss by Fe

Let us assume,

$\begin{array}{rcl}m& =& massoftheelement\\ C_p& =& Heatcapacity\\ △T& =& Changeintemperature\\ T_f& =& Finaltemperature\\ Q& =& Amountofheat\\ & & \end{array}$

Given

Mass of aluminium pellet=5g

Mass of iron pellet=10g

Mass of water= 97.3 g

Specific heat capacityof aluminium pellet=0.89 J ⁰C^-1 g^-1

Specific heat capacityof iron pellet= 0.45 J ⁰C^-1 g^-1

It is known that Specific heat capacityof water =4.18J ⁰C^-1 g^-1

Determine the final temperature

Heat gain by water=Heat loss by Al+ Heat loss by Fe

$\begin{array}{rcl}97.3g{H}_{2}O\times 4.18\frac{J}{g°C}\times \left(T_f-15\right)°C& =& 5.0gAl\times 0.89\frac{J}{g°C}\times \left(100-T_f\right)°+10.0gFe\times 0.45\frac{J}{g°C}\times \left(100-T_f\right)°C\\ 416T_f& =& 9850\\ T_f& =& 23.7°C\end{array}$