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
Q = mc change T. Since they only asked about the heat given off by hot water you can ignore all the stuff about the cold.
Q = .05 (c) (40-9)
HAve to look up specific heat of water.
To calculate the heat given off by the hot water, you can use the equation:
Q = m * c * ΔT
Where:
Q is the amount of heat transferred
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
In this case, you have:
Initial mass of water (m1) = 50g
Initial temperature (T1) = 40°C
Final temperature (T2) = 9°C
Final mass of water (m2) = 50g + 214g = 264g
To calculate the heat transferred, we need to find the change in temperature (ΔT).
ΔT = T2 - T1
ΔT = 9°C - 40°C
ΔT = -31°C
Since the temperature has decreased, the change in temperature is negative.
Now, substitute the values into the equation:
Q = m * c * ΔT
Q = 264g * c * (-31°C)
To solve the equation, we need the specific heat capacity of water (c), which is approximately 4.18 J/g°C.
Substituting this value:
Q = 264g * 4.18 J/g°C * (-31°C)
Calculating the result:
Q = -34092.48 J
Therefore, the heat given off by the hot water is approximately -34092.48 J. The negative sign indicates that heat is being lost by the water.