An open vessel contains 0.5 kg of ice at -20 degree Celsius. The mass of the container can be neglected. Heat is supplied to the vessel at constant rate of 420 J/min for 500 min.
(a) After how many minutes does the ice start to melt?
(b) After how many minutes does the temperature start to rise above 0 degree Celsius?
(c) Plot a curve showing the elapsed time as abscissa and the temperature as ordinate.
To answer these questions, we need to understand the concept of specific heat and phase changes of substances. First, let's break down the steps to find the answers:
(a) To determine when the ice starts to melt, we need to calculate the amount of heat required to raise the temperature of the ice from -20 degree Celsius to 0 degree Celsius. Once this amount of heat is supplied, the remaining heat will be used for the phase change (melting) of the ice.
(b) To find the time at which the temperature starts to rise above 0 degree Celsius, we need to calculate the amount of heat required to raise the temperature of the melted ice (water) from 0 degree Celsius to a higher temperature. The time at which this amount of heat is supplied will be the answer.
(c) To plot the curve, we will use the elapsed time as the x-axis and the temperature as the y-axis. We will calculate the temperature at each elapsed time and plot the data.
Let's start by calculating the heat required for each step:
(a) Heat required to raise the temperature of the ice from -20 degree Celsius to 0 degree Celsius:
The specific heat capacity of ice (c_ice) is 2100 J/(kg·°C). The mass of the ice is 0.5 kg, and the temperature change required is (0 - (-20)) = 20 degree Celsius.
The heat required is given by the formula: Q = m · c · ΔT
Q_1 = (0.5 kg) · (2100 J/(kg·°C)) · (20 °C) = 21,000 J
So, 21,000 J of heat is required to raise the temperature of the ice from -20 degree Celsius to 0 degree Celsius.
(b) Heat required to raise the temperature of the water from 0 degree Celsius to a higher temperature:
The specific heat capacity of water (c_water) is 4200 J/(kg·°C). The mass of the melted ice (water) is still 0.5 kg.
The heat required is given by: Q = m · c · ΔT
Q_2 = (0.5 kg) · (4200 J/(kg·°C)) · ΔT
To find ΔT, we need to know the total heat supplied during the time interval of 500 minutes. We are given that the heat is supplied at a constant rate of 420 J/min for 500 min. Therefore, the total heat supplied is:
Q_total = (420 J/min) · (500 min) = 210,000 J
The remaining heat after melting the ice is:
Q_remain = Q_total - Q_1
Q_remain = 210,000 J - 21,000 J = 189,000 J
Now, we can calculate the temperature rise:
Q_2 = (0.5 kg) · (4200 J/(kg·°C)) · ΔT
189,000 J = (0.5 kg) · (4200 J/(kg·°C)) · ΔT
ΔT = 189,000 J / (0.5 kg · 4200 J/(kg·°C))
ΔT = 90 °C
So, the temperature rises by 90 degree Celsius.
(c) To plot the curve, we need to calculate the temperature at specific time intervals. Let's start with the initial temperature of -20 degree Celsius and plot the points:
Elapsed time: 0 min, Temperature: -20 degree Celsius (initial temperature)
Elapsed time: t, Temperature: 0 degree Celsius (temperature at the start of ice melting)
After that, we will need to calculate the time at which the temperature rises by 90 degree Celsius:
Elapsed time: t + Δt, Temperature: 90 degree Celsius (temperature when the ice has fully melted, and temperature starts to rise)
Now we have all the information to plot the curve. The x-axis will represent the elapsed time, and the y-axis will represent the temperature. We can connect the plotted points with a smooth curve to represent the temperature changes over time.
Note: The curve will be a horizontal line at -20 degree Celsius until point (0, -20). From that point, it will rise linearly until point (t, 0). After that, it will continue to rise linearly until point (t + Δt, 90).