How much heat is required to warm 1.70 l of water from 30.0 ∘c to 100.0 ∘c? (assume a density of 1.0g/ml for the

Heat absorbed is calculated by multiplying the heat capacity of water by mass and the by change in temperature. The heat capacity of water is 4184 J/kg/C.
Thus, heat = mcθ
The mass of water will be given by (1700 × 1)
Thus the heat absorbed will be (1700/1000) × 4184 × 70
                                                 =  497896 J
                                                 = 497.896 kJ

Quantity of heat = mass x specific heat capacity x change in temp. 
Q = m * c * change in Temperature 
Mass = Density * Volume = 1g/mL * 1.7L * 1000mL/1 = 1700g 
Q=1700 * 4.18 * (100.0â’30.0) 
Q = 497420 J

The answer is b metal denting