A 95 gram piece of metal is heated to 98°C and placed into 200mL of water insulated at 20°C. If the equilibrium temperature is 27.20, find:
The heat gained by the water. calories: _________
The specific heat of the metal. cal/(g°C): ________
Hint: 1 mL water has a mass of 1 gram
I have no idea how to begin any of this, much less how to determine the answers for the two parts. Can someone help please?
Anna, I showed you a correct solution, in JOULES. If you want the obsolete unit of calories, then just use the specific heat of water in calories (1 cal/gram per degree C), instead of 4.18 J/gC.
Or, you can convert the answer in Joules to calories (4.18J=1calorie).
I'm still getting the wrong answer. I don't understand this at all. I tried re-posting in the hopes that someone can explain the formula so I can get the correct answer. Thanks very much!
post your work, let me see what you are doing.
6019 wrong
3325 wrong
I can't afford any more wrong answers. Won't get any credit. Thanks again for your help. I'll try posting the question again and see if I can understand the formulas. I appreciate your time.
Well, as you wish, but I suspect if you posted your WORK, it would be solved very quickly what the issue is.
heat = mass * temperature change * specific heat
for the water
heat = 200 g * 7.2 ºC * 1 cal/gºC
heat gained by the water is the heat lost by the metal
for metal
heat = 95 g * 70.8 ºC * s.h.
s.h. = (water heat) / (95 * 70.8)
be aware of significant figures in your calculations and answers
Sure! Let's break this problem down step by step:
Step 1: Calculate the heat gained by the water.
To calculate the heat gained by the water, we can use the formula:
Q = m * c * ΔT
where:
Q = heat gained or lost (in calories)
m = mass of the water (in grams)
c = specific heat capacity of water (in cal/(g°C))
ΔT = change in temperature (in °C)
In this case, we know that the mass of water is 200 mL, and since 1 mL of water is equivalent to 1 gram, the mass of water is also 200 grams.
Given that the initial temperature of the water (T initial) is 20°C, and the final temperature of the water (T final) is 27.20°C, we can calculate the change in temperature (ΔT) as follows:
ΔT = T final - T initial
= 27.20°C - 20°C
= 7.20°C
Now, we can substitute the values into the formula:
Q = 200 g * c * 7.20°C
But we don't have the value of c (specific heat capacity of water) yet, so we will calculate it in the next step.
Step 2: Calculate the specific heat of the metal.
To calculate the specific heat of the metal, we can use the formula:
Q = m * c * ΔT
where:
Q = heat gained or lost (in calories)
m = mass of the metal (in grams)
c = specific heat capacity of the metal (in cal/(g°C))
ΔT = change in temperature (in °C)
We know that the mass of the metal is 95 grams. Also, we assume that there is no heat lost to the surroundings, so the heat gained by the water is equal to the heat lost by the metal. Therefore, we can equate the two equations for heat gained:
200 g * c * 7.20°C = 95 g * c * (98°C - 27.20°C)
Simplifying this equation, we can solve for c, which represents the specific heat capacity of the metal.
Step 3: Calculate the answers for the two parts.
Now that we have the value of c, we can substitute it back into the formula to calculate the heat gained by the water (Q).
Finally, we can calculate the specific heat of the metal (c).
I hope this explanation helps you understand the steps to solve this problem.