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 would be happy to help you with these questions! Let's go through each question and explain how to find the answer.

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 have Q = 1200 calories, m = 200 grams, and ΔT = 3oC. So we can rearrange the formula to solve for c: c = Q / (m * ΔT). Now you can plug in the values and calculate the specific heat capacity.

2) When you pour the hot water into the cup containing the ice, heat will be transferred from the hot water to the ice until they reach the same temperature. This is called heat transfer by conduction. Since the initial temperature of the ice is 0oC and the initial temperature of the water is 80oC, the heat will flow from the hot water to the ice until they both reach the final temperature, which will be somewhere between 0oC and 80oC. Therefore, the answer is 100 g of water at a certain temperature between 0oC and 80oC.

3) To find out how much ice melts, we need to calculate the amount of heat transferred to the ice. 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 have Q = ? (to be determined), m = 20 grams, ΔT = 100oC - 0oC = 100oC (since the iron is heated to 100oC and then dropped on the ice which is at 0oC), and c (specific heat capacity of iron) = 0.108 cal/goC. Now you can plug in the values and calculate the amount of ice that melts.

4) To calculate the amount of heat required to heat the ice cube, 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 have Q = ? (to be determined), m = 50 grams, ΔT = 20oC - (-10oC) = 30oC (since the ice cube is cooled from -10oC to 20oC), and c (specific heat capacity of ice) = 0.5 cal/goC. Now you can plug in the values and calculate the amount of heat required.

I hope this explanation helps you understand how to approach each question and find the answers!