If 11 kJ of heat is transferred to 2300 g of soil at 21oC what would the new temperature of the soil be The specific heat capacity of soil is 0.80 J g oC?

To find the new temperature of the soil, we can use the formula ( Q = mc\Delta T ), where ( Q ) is the heat added (11,000 J), ( m ) is the mass (2300 g), and ( c ) is the specific heat capacity (0.80 J/g°C). Rearranging for ( \Delta T ), we have ( \Delta T = \frac{Q}{mc} = \frac{11,000 , \text{J}}{2300 , \text{g} \times 0.80 , \text{J/g°C}} ). This gives ( \Delta T \approx 6.0°C ), so the new temperature of the soil would be approximately ( 21°C + 6°C = 27°C ).