Calculate the heat given off by the hot water.
GIVEN:
Original quantity of water (hot water) 50g
Orig.temperature :40°C
Final temperature: 9°C
Final quantity of water: 264g
Increase in quantity of water 214g.
PLEASE KINDLY SHOW ME THE EQUATION FOR THIS THANK YOU. <3
To calculate the heat given off by the hot water, we can use the equation:
Q = m * c * ΔT
where:
Q is the heat energy
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
Given:
- Original quantity of water (hot water) = 50g
- Original temperature = 40°C
- Final temperature = 9°C
- Final quantity of water = 264g
- Increase in quantity of water = 214g
Since the quantity of water has increased during the process, we need to consider the change in mass as well. The heat given off by the hot water can be divided into two parts: the heat lost by the original water and the heat lost by the added water. Let's calculate each part separately.
1. Heat lost by the original water:
mass of original water = 50g
change in temperature = (final temperature - initial temperature) = (9°C - 40°C) = -31°C (negative because the temperature decreased)
specific heat capacity of water (c) = 4.18 J/g°C (assuming it is constant)
Q1 = m1 * c * ΔT1
Q1 = 50g * 4.18 J/g°C * (-31°C)
Q1 = -6479 J (Note: negative sign indicates the heat was lost by the water)
2. Heat lost by the added water:
mass of added water = increase in quantity of water = 214g
change in temperature = (final temperature - initial temperature) = (9°C - 0°C) = 9°C
specific heat capacity of water (c) = 4.18 J/g°C (assuming it is constant)
Q2 = m2 * c * ΔT2
Q2 = 214g * 4.18 J/g°C * 9°C
Q2 = 7785.88 J
Total heat given off by the hot water:
Q = Q1 + Q2
Q = -6479 J + 7785.88 J
Q = 1306.88 J
Therefore, the heat given off by the hot water is 1306.88 Joules (J).