Question

one litre of water at 1.00°c is warmed by the addition of 4.00 kj of heat energy. what is the final temperature of the water? (1.00 l of water has a mass of 1.00 kg) t₁ = 1.00°c t₂ =? m = 1.00 kg or 1000g c = 4.19 j/g°c q = 4.00 kj or 4000 j q = mcδt 4000 j = 1000g(4.19 j/g°c)(t₂ - 1.00°c) therefore, the final temperature of the water was 1.95°c.

Explanation:

Step1: Convert heat energy to joules

Given \( Q = 4.00\,\text{kJ}\), since \(1\,\text{kJ}=1000\,\text{J}\), then \(Q = 4000\,\text{J}\).

Step2: Use the heat - transfer formula \(Q = mc\Delta T\)

The specific heat capacity of water \(c = 4.19\,\text{J/g}\cdot^{\circ}\text{C}\), mass \(m = 1000\,\text{g}\) (because \(1.00\,\text{kg}=1000\,\text{g}\)), and \(\Delta T=T_2 - T_1\).
Substitute into the formula: \(4000\,\text{J}=1000\,\text{g}\times4.19\,\text{J/g}\cdot^{\circ}\text{C}\times(T_2 - 1.00^{\circ}\text{C})\).

Step3: Solve for \(T_2\)

First, simplify the equation: \(\frac{4000}{1000\times4.19}=T_2 - 1.00\).
\(\frac{4000}{4190}=T_2 - 1.00\).
\(T_2=1.00+\frac{4000}{4190}\).
\(T_2\approx1.00 + 0.95 = 1.95^{\circ}\text{C}\).

Answer:

\(1.95^{\circ}\text{C}\)