100 mL of 1.00 M HCl is mixed with 100 mL of 1.00 M NaOH in an insulated cup. For neutralisation Delta H = -57.1 kJ mol^-1. Assuming density 1.00 g mL^-1, specific heat 4.18 J g^-1K^-1 and no heat loss to the cup, the temperature rise is

JEE Main levelSingle correct optionDifficulty 2 of 3td085

$100 mL$ of $1.00 M$ HCl is mixed with $100 mL$ of $1.00 M$ NaOH in an insulated cup. For neutralisation $Δ H = - 57.1 kJ mo l^{- 1}$ . Assuming density $1.00 g m L^{- 1}$ , specific heat $4.18 J g^{- 1} K^{- 1}$ and no heat loss to the cup, the temperature rise is

Moles of water formed

= 0.100 \times 1.00 = 0.100 mol

, so

q = 0.100 \times 57100 = 5710 J

. The mass heated is

200 g

, so

Δ T = 5710/ (200 \times 4.18) = 6.83 K