Calculate the amount of heat required to convert 155 g of ice at -10oC to steam at 110oC?
(155g)x[0-(-10)]C x (.50cal/g x C) = 775cal
(155g) x (80cal/g) = 12400cal
(155g)x(100-0)C x (1.00cal/g x C) = 15500cal
(155g) x (540cal/g) = 83700cal
775cal + 12400cal + 15500cal +83700cal = 112375cal
Is that correct?
I don't know what the C stands for but you seem to have ignored it in the calculation. I didn't go through the calculations with a calculator but the methods seems OK EXCEPT that you haven't added in the heat from steam at 100 to steam at 110
Yes, your calculation is correct. To calculate the amount of heat required to convert a given mass of ice at a certain temperature to steam at a different temperature, you need to consider the following steps:
1. Calculate the heat required to raise the temperature of the ice from -10°C to the melting point of ice (0°C). This is done using the formula: Q = mass × temperature difference × specific heat capacity of ice.
In this case: Q1 = (155 g) × (0 - (-10))°C × (0.50 cal/g°C) = 775 cal.
2. Calculate the heat required to melt the ice at 0°C. This is given by: Q2 = mass × heat of fusion of water.
In this case: Q2 = (155 g) × (80 cal/g) = 12,400 cal.
3. Calculate the heat required to raise the temperature of the melted ice (now water) from 0°C to the boiling point of water (100°C). Using the same formula as step 1, we have: Q3 = (155 g) × (100 - 0)°C × (1.00 cal/g°C) = 15,500 cal.
4. Calculate the heat required to convert the water at 100°C to steam at 100°C. This is given by: Q4 = mass × heat of vaporization of water.
In this case: Q4 = (155 g) × (540 cal/g) = 83,700 cal.
Finally, sum up all the quantities of heat required in each step: Q_total = Q1 + Q2 + Q3 + Q4 = 775 cal + 12,400 cal + 15,500 cal + 83,700 cal = 112,375 cal.
So, your answer is correct. The amount of heat required to convert 155 g of ice at -10°C to steam at 110°C is 112,375 calories.