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 ${75}^{o}C$. $T$ is equal to
(Given : Room temperature $={30}^{o}C$, specific heat of copper $=0.1cal/g{m}^{o}C$)

• A. ${825}^{o}C$
• B. ${800}^{o}C$
• C. ${885}^{o}C$
• D. ${1250}^{o}C$

Answer 1

The correct option is C ${885}^{o}C$

The amount of heat lost by copper ball $=m_{ball}{C}_{Cu}\Delta T_{ball}$
$=\left(100gm\right)\left(0.1cal/gm˚C\right)(T-75)˚C$
The amount of heat gained by calorimeter
$=m_{calorie}{C}_{calorie}\Delta T_{calorie}$
$=\left(100gm\right)\left(0.1cal/gm˚C\right)(75-30)˚C$
Amount of heat gained by water $=m_w{C}_w\Delta T_w$
$=\left(170gm\right)\left(1cal/gm˚C\right)(75-30)˚C$
Hence, from conservation of energy,
$100\times 0.1\times (T-75)=100\times 0.1\times (75-30)$
$+170\times 1\times (75-30)$
$\Rightarrow T=885˚C$