Question

A copper ball of mass $100gm$ is at a temperature $T$. It is dropped in a copper calorimeter of mass $100gm$, filled with $170gm$ of water at room temperature. Subsequently, the temperature of the system is found to be $75°C$. $T$ is given by : (Given : room temperature $=30°C$, specific heat of copper $=0.1cal/gm°C$)

• A. $800°C$
• B. $885°C$
• C. $1250°C$
• D. $825°C$

Answer 1

The correct option is B

$885°C$

Step 1. Given data:

$m_{cal}=100gm$, Mass of copper calorimeter

$m_c=100gm$, Mass of copper ball

$m_w=170gm$, Mass of water

$C_c=0.1cal/gm°C$ specific heat of copper

$C_{cal}=0.1cal/gm°C$ specific heat of copper calorimeter,

$C_w=1cal/gm°C$ specific heat of water,

$T\text{'}=75°C$ Final temperature of copper ball,

$T_{cal}=T_w=30°C$ Initial temperature of water and copper calorimeter

$T{\text{'}}_{cal}=T{\text{'}}_w=75°C$ Final temperature of water and copper calorimeter

Initial temperature of copper ball, $T$

Step 2. Calculating initial temperature of copper ball, $T$

Applying conservation of energy,

$Heatlostbycopperball=\left(Heatgainedbycoppercalorimeter\right)+\left(Heatgainedbywater\right)$

$\begin{array}{rcl}& \Rightarrow & m_c{C}_c\left(T-T\text{'}\right)=m_{cal}{C}_{cal}\left(T{\text{'}}_{cal}-T_{cal}\right)+m_w{C}_w\left(T{\text{'}}_w-T_w\right)\\ & \Rightarrow & 100\times \left(0.1\right)\times \left(T-75\right)=100\times \left(0.1\right)\times \left(75-30\right)+170\times 1\times \left(75-30\right)\\ & \Rightarrow & 10T-750=450+7650\\ & \Rightarrow & 10T=8850\\ \therefore T& =& 885°C\end{array}$

Thus, $T$ is given by $885°C$.

Hence, correct option is (B).