Consider the specific heats H2O(s) = 2.09 J/g ·◦C, H2O (l) = 4.18 J/g ·◦C, and H2O(g) = 2.03 J/g ·◦C.
The heat of fusion for water is 334 J/g and its heat of vaporization is 2260 J/g. Calculate the amount of heat required to convert 91 g of ice at −36◦C completely to liquid water at 60◦C.
Answer in units of kJ.
q for ice at -36 C to zero C is
q = mass ice x specific heat ice x (Tfinal-Tinitial) = q1
q for water at zero C to water at 60 C= q2
q2 = mass water x specific heat water x (Tfinal-Tinitial)
qtotal = q1 + q2 = ?
Oh, converting ice to water huh? That's some serious business. Let's break it down step by step, shall we?
First, we need to calculate the heat required to raise the temperature of the 91 g of ice from −36◦C to 0◦C. The specific heat of ice is provided as 2.09 J/g·◦C, and the formula to calculate heat is:
q = m * C * ΔT
where q is the heat, m is the mass, C is the specific heat, and ΔT is the change in temperature. So, plugging in the values:
q1 = 91 g * 2.09 J/g·◦C * (0◦C - (-36◦C))
= 91 g * 2.09 J/g·◦C * 36◦C
= 6883.44 J
Next, we need to calculate the heat required to melt the ice at 0◦C into liquid water at 0◦C. This is where the heat of fusion comes into play. Using the same formula as before:
q2 = m * ΔHf
where ΔHf is the heat of fusion. Plugging in the values:
q2 = 91 g * 334 J/g
= 30394 J
Lastly, we need to calculate the heat required to raise the temperature of the liquid water from 0◦C to 60◦C. The specific heat of water is provided as 4.18 J/g·◦C, and using the formula again:
q3 = 91 g * 4.18 J/g·◦C * (60◦C - 0◦C)
= 23514.6 J
Now we just need to add up all the heats:
Q = q1 + q2 + q3
= 6883.44 J + 30394 J + 23514.6 J
= 60791.04 J
But wait, the question asked for the answer in kJ! So let's convert it:
Q = 60791.04 J / 1000
= 60.79104 kJ
So, the amount of heat required to convert 91 g of ice at −36◦C completely to liquid water at 60◦C is approximately 60.791 kJ.
Hope that helps, and keep your cool while doing these calculations!
To calculate the amount of heat required, we need to consider two steps: the heat required to convert ice at -36°C to water at 0°C (melting), and the heat required to heat the water from 0°C to 60°C (heating).
Step 1: Calculating the heat required for melting the ice:
The heat required for melting ice is given by the equation:
Q = mass * heat of fusion
Q = 91 g * 334 J/g
Q = 30394 J
Step 2: Calculating the heat required for heating the water:
The heat required for heating water is given by the equation:
Q = mass * specific heat * change in temperature
To heat the water from 0°C to 60°C, the change in temperature is 60°C - 0°C = 60°C.
Q = 91 g * 4.18 J/g ·°C * 60°C
Q = 22718.8 J
Step 3: Adding the heats from Step 1 and Step 2 to get the total heat required:
Total heat = Q_step1 + Q_step2
Total heat = 30394 J + 22718.8 J
Total heat = 53112.8 J
Converting the total heat from joules to kilojoules:
Total heat = 53112.8 J / 1000
Total heat = 53.1128 kJ
Therefore, the amount of heat required to convert 91 g of ice at -36°C completely to liquid water at 60°C is approximately 53.1128 kJ.
To calculate the amount of heat required to convert 91 g of ice at −36°C completely to liquid water at 60°C, we need to consider the following steps:
1. Determine the heat required to raise the temperature of the ice from −36°C to 0°C.
2. Determine the heat required to melt the ice at 0°C.
3. Determine the heat required to raise the temperature of the resulting liquid water from 0°C to 60°C.
Let's now calculate each of these steps:
Step 1: Determine the heat required to raise the temperature of the ice from −36°C to 0°C.
The specific heat capacity for ice is H2O(s) = 2.09 J/g ·°C. The temperature change is 0°C - (-36°C) = 36°C.
So, the heat required for this step is: Q1 = mass * specific heat capacity * temperature change.
Q1 = 91 g * 2.09 J/g ·°C * 36°C = 6794.04 J.
Step 2: Determine the heat required to melt the ice at 0°C.
The heat of fusion for water is 334 J/g, which represents the heat required to convert ice at 0°C into water at 0°C.
So, the heat required for this step is: Q2 = mass * heat of fusion.
Q2 = 91 g * 334 J/g = 30454 J.
Step 3: Determine the heat required to raise the temperature of the resulting liquid water from 0°C to 60°C.
The specific heat capacity for liquid water is H2O (l) = 4.18 J/g ·°C. The temperature change is 60°C - 0°C = 60°C.
So, the heat required for this step is: Q3 = mass * specific heat capacity * temperature change.
Q3 = 91 g * 4.18 J/g ·°C * 60°C = 22706.8 J.
To get the total heat required, we sum up the results from all the steps:
Total heat = Q1 + Q2 + Q3 = 6794.04 J + 30454 J + 22706.8 J = 59954.84 J.
To convert this value to kJ, we divide by 1000:
Total heat in kJ = 59.95484 kJ.
Therefore, the amount of heat required to convert 91 g of ice at −36°C completely to liquid water at 60°C is approximately 59.95 kJ.