QUESTION 1: How many Joules of must be removed from 220 g of water to cool it fr
ID: 963826 • Letter: Q
Question
QUESTION 1:
How many Joules of must be removed from 220 g of water to cool it from 87 oC to 10 oC if water has a specific heat of 4.184 J/(g °C)?
Question 2: A 110 g sample of copper with a specific heat of .02 J/(g °C) is heated to 82.4 °C and then placed in a container of water at 22.3 °C. The final temperature of the water and copper is 24.9 °C. What is the mass of the water in the container, assuming that all of the heat lost by the copper is gained by the water?
can someone answer these two questions for me by working the problem out? I have to have them for a calorimetry report and she wont except the answer unless work is shown! thank you!
Explanation / Answer
Q.NO :1
Q relaeased = m*s*dT
= 220*4.18*(87-10) = 70.8 kj
Q.no : 2
q lost by copper = q gained by water
mcopper*s*DT = mwater*s*DT
110*0.02*(82.4-24.9) = x*4.18*(24.9-22.3)
x= mass of water = 11.64 grams