Answer
$70.5^{\circ}C$
Work Step by Step
We know that
$T=\frac{m_mC_mT_m+m_{cr}C_{cr}T_{cr}+m_cC_cT_c}{m_mC_m+m_{cr}C_{cr}+m_cC_c}$
We plug in the known values to obtain:
$T=\frac{(0.116Kg)(1090J/Kg.K)(297K)+(0.0122Kg)(4186J/Kg.K)(278K)+(0.225Kg)(4186J/Kg.K)(353.3K)}{(0.116Kg)(1090J/Kg.K)+(0.0122Kg)(4186J/Kg.K)+(0.225Kg)(4186J/Kg.K)}$
$\implies T=343.5K=(343.5-273)C^{\circ}=70.5^{\circ}C$
