Answer is: 6,16 kJ.
1) changing temperature of ice from -25°C to 0°C.
Q₁ = m·C·ΔT
Q₁ = 18 g · 2 J/g·°C · 25°C
Q₁ = 900 J.
m(H₂O) = 1mol · 18 g/mol = 18 g.
C - <span>specific heat of ice.
</span>2) changing temperature of water from 0°C to 70°C.
Q₁ = m·C·ΔT
Q₁ = 18 g · 4,18 J/g·°C · 70°C
Q₁ = 5266,8 J.
C - specific heat of water.
Q = Q₁ + Q₂ = 900 J + 5266,8 J
Q = 6166,8 J = 6,16 kJ.
A calorimeter experiment is a set-up that provides insulation so that no heat escapes to the surroundings and all energy can be accounted for. It can be done at either constant volume or constant pressure. So, the answer to this is knowing the mass of water, the specific heat which is an empirical data, and the change in temperature which can be measured using a thermometer. This experiment could measure the mass of an unknown substance added or the specific heat of the substance or the calorimeter. <em>The answer is D.</em>
