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 can help you with these questions. Let's go through each one step by step.
Question 1: A 200-gram chunk of stuff rises in temperature by 3oC when you input 1200 calories of heat. What is its specific heat capacity?
To find the specific heat capacity, we can use the formula:
Heat (in calories) = Mass (in grams) × Specific heat capacity (in cal/g°C) × Temperature change (in °C)
From the given information, we can plug the values into the formula:
1200 calories = 200 grams × Specific heat capacity × 3°C
Now, solve for the specific heat capacity:
Specific heat capacity = 1200 calories / (200 grams × 3°C)
Specific heat capacity = 2 cal/goC
Therefore, the correct answer is B) 2 cal/goC.
Question 2: 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?
When you pour the hot water into the cup containing the ice, the heat from the water will transfer to the ice, causing it to melt.
The final equilibrium temperature will be at 0oC, so all the ice will melt. However, not all the hot water will completely cool down, since it has a higher initial temperature.
Therefore, the correct answer is A) 100 g of 0oC water and 100 g of 80oC water.
Question 3: 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.
To find the amount of ice that melts, we can use the formula:
Heat gained by ice (in calories) = Mass of ice (in grams) × Specific heat capacity of ice (in cal/goC) × Temperature change (in °C)
From the given information, the iron transfers heat to the ice until it reaches the melting point of the ice, which is 0oC.
The heat gained by the ice is equal to the heat lost by the iron:
Heat gained by ice = Heat lost by iron
Therefore, we can set up the equation:
Mass of ice × Specific heat capacity of ice × (0°C - (-100°C)) = Mass of iron × Specific heat capacity of iron × (0°C - 100°C)
Now, plug in the values:
Mass of ice × 0.5 cal/goC × 0°C - (-100°C) = 20 g × 0.108 cal/goC × (0°C - 100°C)
Mass of ice × 0.5 cal/goC × 100°C = 20 g × 0.108 cal/goC × (-100°C)
Solve for the mass of ice:
Mass of ice = (20 g × 0.108 cal/goC × (-100°C)) / (0.5 cal/goC × 100°C)
Mass of ice = 2.16 g
Therefore, the correct answer is B) 2.7 g.
Question 4: 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.)
To find the heat required to heat the ice cube, we can use the formula:
Heat (in calories) = Mass (in grams) × Specific heat capacity (in cal/g°C) × Temperature change (in °C)
From the given information, we need to calculate the heat required to heat the ice cube from -10oC to 0oC and then from 0oC to 20oC.
Heat required = (50 g × 0.5 cal/goC × (0°C - (-10°C))) + (50 g × 1 cal/goC × (20°C - 0°C))
Heat required = (50 g × 0.5 cal/goC × 10°C) + (50 g × 1 cal/goC × 20°C)
Heat required = 250 cal + 1000 cal
Heat required = 1250 cal
Therefore, the correct answer is A) 1250 cal.
Sure! I'd be happy to help you with those questions. Let's go through each question one by one.
Question 1:
To find the specific heat capacity, we can use the formula:
Q = mcΔT
where Q is the amount of heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
In this case, we know that the mass (m) is 200 grams, the change in temperature (ΔT) is 3oC, and the heat input (Q) is 1200 calories.
So, plugging in the values:
Q = mcΔT
1200 cal = 200 g * c * 3oC
To solve for c, we can rearrange the equation:
c = Q / (m * ΔT)
c = 1200 cal / (200 g * 3oC)
Now, you can calculate the value of c and find the specific heat capacity.
Question 2:
When you pour hot water into the cup containing ice, heat will be transferred from the hot water to the ice. The heat transfer will cause the ice to melt and the water to cool down.
So, after pouring the hot water into the cup, you will wind up with a mixture of water and ice at some temperature between 0oC and 80oC. To determine the exact answer, you'll need to calculate the final temperature of the mixture using the principles of heat transfer.
Question 3:
To find out how much ice melts when a heated piece of iron is dropped onto it, we need to consider the heat transferred from the iron to the ice. The heat transferred can be calculated using the formula:
Q = mcΔT
In this case, the mass of the iron (m) is given as 20g, the specific heat capacity of iron (c) is given as 0.108 cal/goC, and the change in temperature (ΔT) is 100oC (since the iron is heated to 100oC). The heat transferred (Q) will be used to melt the ice.
To figure out how much ice melts, we need to know the heat required to melt 1g of ice, which is 80 calories. We can then calculate how many grams of ice can be melted using the heat transferred (Q) from the iron.
Question 4:
To find the amount of heat required to heat a 50g ice cube to become liquid water at 20oC, we need to consider the heat transfer involved at each stage. First, the ice cube needs to be heated from -10oC to 0oC, and then it needs to melt at 0oC, and finally, it needs to be heated from 0oC to 20oC.
Using the formula Q = mcΔT, we can calculate the heat required for each stage and sum them up to get the total heat required.
To find the specific heat of ice, we are given that it is 0.5 cal/goC. The mass (m) of the ice cube is 50g, the change in temperature (ΔT) for the first stage is 10oC (from -10oC to 0oC), and for the third stage is 20oC (from 0oC to 20oC).
By adding up the heat required for each stage, we can find the total heat required to heat the ice cube until it becomes liquid water at 20oC.
I hope this helps! Let me know if you have any further questions.