Hey There, I have 4 questions that I am totally stumped on... Can anyone help?
A 200-gram chunk of stuff rises in temperature by 3oC when you input 1200 calories of heat. What is its specific heat capacity?
Answers
A)50 cal/goC.
B) 2 cal/goC.
C) 5 cal/goC.
D) 0.02 cal/goC.
You have two cups, one containing 100 g of ice at 0oC, the other containing 100 g of water at 80oC. You pour the hot water into the cup containing the ice. What do you wind up with?
Answers
A)100 g of 0oC water and 100 g of 80oC water.
B)50 g of 0oC ice and 50g of 0oC water.
C)200 g of 0oC water.
D)100 g of 0oC water.
A 20-g piece of iron is heated to 100oC. It is then dropped on a very large chunk of ice. How much ice melts? The specific heat capacity of iron is 0.108 cal/goC.
Answers
A) 1.35 g
B) 2.7 g
C) 5.4 g
D) 20 g
A 50 gram ice cube is cooled to -10oC in the freezer. How many calories of heat are required to heat it until it becomes liquid water at 20oC? (The specific heat of ice is 0.5 cal/goC.)
Answers:
A)1250 cal.
B) 4000 cal.
C) 5000 cal.
D) 5250 cal.
Sure! I'd be happy to help you with these questions. Let's go through each question one by one and find the answers.
Question 1:
To find the specific heat capacity, we can use the formula:
Heat (in calories) = mass (in grams) × specific heat capacity (in cal/goC) × temperature change (in oC)
Given:
Mass = 200 grams
Heat = 1200 calories
Temperature change = 3oC
Let's rearrange the formula to solve for the specific heat capacity:
Specific heat capacity = Heat / (mass × temperature change)
Plugging in the values, we get:
Specific heat capacity = 1200 calories / (200 grams × 3oC)
Simplifying the calculation, we find:
Specific heat capacity = 2 cal/goC
So, the answer is B) 2 cal/goC.
Question 2:
When you pour the hot water into the cup containing the ice, the heat from the hot water will transfer to the ice, causing it to melt and mix with the water.
Since the mass of ice and water in both cups is the same (100 grams), and the final temperature will be a combination of the initial temperatures (0oC for ice and 80oC for water), we can conclude that the answer is A) 100 g of 0oC water and 100 g of 80oC water.
Question 3:
To determine how much ice melts when the heated iron is dropped on it, we need to calculate the heat that is transferred from the iron to the ice.
Using the formula:
Heat (in calories) = mass (in grams) × specific heat capacity (in cal/goC) × temperature change (in oC)
Given:
Mass of iron = 20 grams
Specific heat capacity of iron = 0.108 cal/goC
Temperature change of iron = 100oC
Let's calculate the heat transferred from the iron to the ice:
Heat = mass × specific heat capacity × temperature change
Heat = 20 grams × 0.108 cal/goC × 100oC
Simplifying the calculation, we find:
Heat = 216 calories
Since 1 gram of ice requires 80 calories of heat to melt, we can find the amount of ice melted:
Ice melted = Heat / 80 calories
Ice melted = 216 calories / 80 calories
Simplifying the calculation, we find:
Ice melted = 2.7 grams
So, the answer is B) 2.7 g.
Question 4:
To find the amount of heat required to heat the ice cube to become liquid water at 20oC, we need to consider two steps: first, raising the temperature of the ice from -10oC to 0oC, and then melting the ice at 0oC.
Using the formula for temperature change:
Heat (in calories) = mass (in grams) × specific heat capacity (in cal/goC) × temperature change (in oC)
Given:
Mass of ice = 50 grams
Specific heat capacity of ice = 0.5 cal/goC
Temperature change to raise from -10oC to 0oC = 10oC
First, let's calculate the heat required to raise the temperature of the ice:
Heat = mass × specific heat capacity × temperature change
Heat = 50 grams × 0.5 cal/goC × 10oC
Simplifying the calculation, we find:
Heat = 250 calories
Next, we need to calculate the heat required to melt the ice:
Heat = mass × heat of fusion
Heat = 50 grams × 80 calories/gram
Simplifying the calculation, we find:
Heat = 4000 calories
Now, let's add both heats together to find the total heat required:
Total heat = Heat to raise temperature + Heat to melt ice
Total heat = 250 calories + 4000 calories
Simplifying the calculation, we find:
Total heat = 4250 calories
So, the answer is D) 4250 cal.
I hope this helps you with your questions! Let me know if there's anything else I can assist you with.