Question

A calorimeter of water equivalent $5\times {10}^{-3}kg$ contains $25\times {10}^{-3}kg$ of water. It takes 3 minutes to cool from ${28}^{\circ }C$ to ${21}^{\circ }C$. When the same calorimeter is filled with $30\times {10}^{-3}kg$ of turpentine oil, then it takes 2 minutes to cool from ${28}^{\circ }C$ to ${21}^{\circ }C$. Find the specific heat of turpentine oil.

• A. $1.5Jk{g}^{-1}\circ C^{-1}$
• B. $0.5Jk{g}^{-1}\circ C^{-1}$
• C. $45Jk{g}^{-1}\circ C^{-1}$
• D. $5Jk{g}^{-1}\circ C^{-1}$

Answer 1

The correct option is B $0.5Jk{g}^{-1}\circ C^{-1}$

$\frac{(m+w)\theta }{t_1}=\frac{(m_t{c}_t+w)\theta }{t_2}=\frac{(25\times {10}^{-3}+5\times {10}^{-3})\times 7}{3}=\frac{(30\times {10}^{-3}{c}_t+5\times {10}^{-3})\times 7}{2}{c}_t=0.5Jk{g}^{-1}\circ C^{-1}$