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).
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?
6 answers
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.
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
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
2 answers