"Find the total energy lost by the contents of a Styrofoam picnic cooler when a 3.60 kg block of ice placed inside the cooler completely melts (at 0 Celsius). The ice had an initial temperature of -10.00 Celsius." (Answer is 1.27*10^3 kJ).
I was thinking of using q=mct, but I didn't get the answer. I might have done something wrong though...
The ice loses heat in moving from a temperature of -10.00 C to O C. For that you use q = mc*delta T. Remember to use the specific heat of ICE (not water) for c in the equation. Then the ice melts and that is mass ice x heat fusion. Add the two qs together. That should do it.
Find the total energy
To find the total energy lost by the contents of the Styrofoam picnic cooler, we need to calculate the heat transfer required to change the temperature of the ice from -10.00 Celsius to 0 Celsius and then to melt it completely at 0 Celsius. Let's break down the steps to solve this problem using the equation q = mcΔT.
Step 1: Calculate the heat transfer to raise the temperature of the ice from -10.00 Celsius to 0 Celsius:
Given:
mass of ice (m) = 3.60 kg
specific heat capacity of ice (c) = 2.09 J/g°C (or 2090 J/kg°C)
temperature change (ΔT) = 0°C - (-10.00°C) = 10.00°C
Use the formula: q = mcΔT
q1 = (mass of ice) * (specific heat capacity of ice) * (temperature change)
= (3.60 kg) * (2090 J/kg°C) * (10.00°C)
= 75,240 J
Step 2: Calculate the heat transfer to melt the ice at 0 Celsius:
Given:
latent heat of fusion for ice (L) = 334 kJ/kg (or 334,000 J/kg)
Use the formula: q = mL
q2 = (mass of ice) * (latent heat of fusion)
= (3.60 kg) * (334,000 J/kg)
= 1,202,400 J
Step 3: Calculate the total energy lost by the contents of the picnic cooler:
The total energy lost is the sum of the heat transfer at each step:
total energy lost = q1 + q2
= 75,240 J + 1,202,400 J
= 1,277,640 J
Convert the answer to kJ by dividing by 1000:
total energy lost = 1,277,640 J / 1000
= 1277.64 kJ
≈ 1.28 x 10^3 kJ (rounded to three significant figures)
Therefore, the total energy lost by the contents of the Styrofoam picnic cooler is approximately 1.28 x 10^3 kJ (or 1.27 x 10^3 kJ, as given in the answer).