Question

The heat produced on complete combustion of $10$ kg of fuel could raise the temperature of $2$ kg of water from ${20}^o$ to ${70}^o$ Calculate the calorific value.

• A. $4.186\times {10}^{-3}$ J/kg
• B. $4.186\times {10}^{-2}$ J/kg
• C. $4.186\times {10}^{-4}$ J/kg
• D. $4.186\times {10}^{-8}$ J/kg

Answer 1

The correct option is B $4.186\times {10}^{-2}$ J/kg

We know

Mass of fuel$\times$ calorific value=mass of water $\times$specific heat $\times$temperature difference

Given mass of fuel $m_f$=10 kg and mass of water $m^{}$ = 2kg

$\Delta T$=${70}^0-{20}^0={50}^{0}C$

Specific heat of water =$4.186J/{grams}^{0}C=4.186\times {10}^{-3}J/{kg}^{0}C$

Now putting the above values in the equation we get

$C_V=\frac{2\times 4.186\times {10}^{-3}\times (70-20)}{10}=4.186\times {10}^{-2}$ J/kg