Question 16A sample of nickel metal is heated to 210.25 °C then dropped into 125.0 grams of 5.25 °C liquidwater. Heat exchanges between the nickel and water until the final temperature of 25.00 °C isreached. (cs Ni = 0.440 J/g °C and cs.H20 = 4.184 J/g °C)%3DCalculate the following (be sure to include correct significant figures AND sign):Change in Temperature of Nickel, in °C:°CChange in Temperature of Water, in °C:°CHeat absorbed by the water, in Joules (J):Mass of Nickel in grams (g):Next

A heated nickel sample is dropped into water.Change in temperature of nickel=25.00-210.25=-185.25°CChange in temperature of water=25.00-5.25=19.75°CMass of water (m)=125.0 gSpecific heat of water (c)=4.184 J/g°CHeat absorbed by water, q=m × c × change in temperature of water=125.0×4.184×19.75=10,329.25 J≃10330 JThe heat absorbed by water will be the heat lost by nickel. This can be used to calculate the mass of the nickel sample. Heat lost by nickel (Ni) sample, q=-10329.25 JSpecific heat capacity of nickel (c)=0.440 J/g°CMass of Ni sample (m)=qc×change in temperature of Ni=-10329.250.440×(-185.25)=126.7 g Therefore, Change in temperature of Nickel, in ⁰C: -185.25 ⁰C Change in temperature of water, in ⁰C : 19.75 ⁰C Heat absorbed by water, in Joules (J) : 10330 J Mass of Nickel in grams (g) : 126.7 g

Science

Chemistry

A sample of nickel metal is heated to 210.25 °C then dropped into 125.0 grams of 5.25 °C liquid water. Heat exchanges between the nickel and water until the final temperature of 25.00 °C is reached. (cs Ni = 0.440 J/g °C and cs.H20 = 4.184 J/g °C) %3! %3D Calculate the following (be sure to include correct significant figures AND sign): °C Change in Temperature of Nickel, in °C: °C Change in Temperature of Water, in °C: Heat absorbed by the water, in Joules (J): Mass of Nickel in grams (g):

A sample of nickel metal is heated to 210.25 °C then dropped into 125.0 grams of 5.25 °C liquid water. Heat exchanges between the nickel and water until the final temperature of 25.00 °C is reached. (cs Ni = 0.440 J/g °C and cs.H20 = 4.184 J/g °C) %3! %3D Calculate the following (be sure to include correct significant figures AND sign): °C Change in Temperature of Nickel, in °C: °C Change in Temperature of Water, in °C: Heat absorbed by the water, in Joules (J): Mass of Nickel in grams (g):

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

How much heat, in joules (J), must be added to a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C?
The specific heat of toluene (C7H8), is 1.13 J/g•K. How many kJ of heat are necessary to raise the temperature of 65.3 g of toluene from 19.3 °C to 30.4 °C?
2.Can you help me solve this
How much energy is absorbed when 5.00 g of H2O(l) is heated from 35.0 C to 65.0 C? The specific heat of water, Cs = 4.184 J/gC. Report your answer to three significant figures without any units.
4. Aluminum metal melts at 658.5°C. What would be the final temperature if 28.0 grams of solid aluminum at this temperature were placed into 520 grams of liquid aluminum at 1000.0°C? The heat of fusion for aluminum is 396 J/ -1 -1 g, and the specific heat of liquid aluminum is 0.481 J-g`¹deg-¹. (Note: this is the aluminum analog to the process you just did in lab!)
IT The temperature of a sample of aluminum increased by 24.6 °C when 255 J of heat was applied. Substance Specific heat J/(g lead 0.128 What is the mass of the sample? silver 0.235 copper 0.385 iron 0.449 aluminum 0.903 m =
Find the final temperature Suppose we mix 90.0 grams of hot water (90.0°C) with 10.0 grams of cold water (10.0°C). Let x = the final temperature. C=4.184 J/g.°C a. Set up an expression for the energy released (q) by the hot water (Aqhot motCAThot) b. Set up an expression for the energy absorbed (q) by the cold water (Aqcold=mcold CATcold) c. Calculate Tf
A 5.0 g sample of water starting at 59.4°C loses 439 J of energy in the form of heat. What is the final temperature of the water after this heat loss? The specific heat for water is 4.18 J/(g.°C). Temperature = °C
A hot 122.8 g lump of an unknown substance initially at 154.8 °C is placed in 35.0 mL of water initially at 25.0 °C and the system is allowed to reach thermal equilibrium. The final temperature of the system is 56.9 °C. Using this information and the specific heat values for several metals in the table, identify the unknown substance. Assume no heat is lost to the surroundings. graphite zinc aluminum titanium tungsten rhodium Substance Specific heat (J/(g-°C)) aluminum 0.897 graphite 0.709 rhodium 0.243 titanium 0.523 0.132 0.388 4.184 tungsten zinc water
When a 45.0 g sample of an alloy at 100.0 °C is dropped into 100.0 g of water at 25.0 °C, the final temperature is 37.0 °C. What is the specific heat of the alloy? The specific heat of H20(1) is 4.184 J'g-1.°c-1. q = mc AT O A. 0.423 J-g-1.°c-1 O B. 9.88 J-g-1.°C-1 O C. 48.8 J'g-1.°C-1 O D. 1.77 Jg-1.°c-1
The specific heat capacity of liquid mercury is 0.14 J/g-K. How many kilojoules of heat are needed to raise the temperature of 9.00 g of mercury from 25.1 °C to 65.3 °C?
The specific heat of substance C is 0.93 J/(g K) and that of substance D is 1.8 J/(g K). You are given an unknown that could be pure substance C, pure substance D, or a homogeneous mixture of C and D. In the lab you determine that it requires 23.3 J of heat energy to raise the temperature of a 25.0 g sample of the unknown by 1.0 K. What conclusions can you make about the identity of your unknown, from this data?