Part A A 11.5-g sample of granite initially at 83.0 °C is immersed into 29.0 g...

Question

Question

Part A A 11.5-g sample of granite initially at 83.0 °C is immersed into 29.0 g of water initially at 24.0 °C. What is the fin

Answer 1

Answer #1

m(water) = 29.0 g

T(water) = 24.0 oC

C(water) = 4.18 J/goC

m(granite) = 11.5 g

T(granite) = 83.0 oC

C(granite) = 0.79 J/goC

T = to be calculated

Let the final temperature be T oC

use:

heat lost by granite = heat gained by water

m(granite)*C(granite)*(T(granite)-T) = m(water)*C(water)*(T-T(water))

11.5*0.79*(83.0-T) = 29.0*4.18*(T-24.0)

9.085*(83.0-T) = 121.22*(T-24.0)

754.055 - 9.085*T = 121.22*T - 2909.28

T= 28.1 oC

Answer: 28.1 oC

Answer 2

Similar Homework Help Questions

Part A A 12.0 x sample of granite initially at 83.0 is immersed into 250 g...

Part A A 12.0 x sample of granite initially at 83.0 is immersed into 250 g of water initially at 25.0°C. What is the final temperature of both substances when they reach thermal equilibrium? (For water, C. = 4.18J/g and for granite, C, = 0.790J/gºC.) Express your answer using three significant figures. AED + O ? 19.21 Submit Previous Answers Request Answer * Incorrect; Try Again; 5 attempts remaining
A 12.5 g sample of granite initially at 82.0oC is immersed into a 25.0 g of...

A 12.5 g sample of granite initially at 82.0oC is immersed into a 25.0 g of water initially at 22.0oC. What is the final temperature of both substances when they reach thermal equilibrium? (For water, c = 4.184 J/g.oC, and for granite, c = 0.790 J/g.oC.)
A 296.0 g piece of granite, heated to 601.0°C in a campfire, is dropped into 1.10...

A 296.0 g piece of granite, heated to 601.0°C in a campfire, is dropped into 1.10 L water (d = 1.00 g/mL) at 25.0°C. The molar heat capacity of water is cp,water = 75.3 J/(mol ·°C), and the specific heat of granite is cs,granite = 0.790 J/(g ·°C). Calculate the final temperature of the granite.

A 301.0 g piece of granite, heated to 561.0°C in a campfire, is dropped into 1.00...

A 301.0 g piece of granite, heated to 561.0°C in a campfire, is dropped into 1.00 L water (d = 1.00 g/mL) at 25.0°C. The molar heat capacity of water is cp,water = 75.3 J/(mol ·°C), and the specific heat of granite is cs,granite = 0.790 J/(g ·°C). Calculate the final temperature of the granite.
A piece of copper metal is initially at 83.0°C. It is dropped into a coffee cup...

A piece of copper metal is initially at 83.0°C. It is dropped into a coffee cup calorimeter containing 30.0 9 of water at a temperature of 10.0°c. After stirring, the final temperature of both copper and water is 25.0°c. Assuming no heat losses, and that the specific heat (capacity) of water is 4.18 J/(g.), what is the heat capacity of the copper in J/K?
1. A 31.6 g wafer of pure gold initially at 69.4 ∘C is submerged into 63.8...

1. A 31.6 g wafer of pure gold initially at 69.4 ∘C is submerged into 63.8 g of water at 27.7 ∘C in an insulated container. What is the final temperature of both substances at thermal equilibrium? 2.Two substances, A and B, initially at different temperatures, come into contact and reach thermal equilibrium. The mass of substance A is 6.04 gand its initial temperature is 21.0 ∘C . The mass of substance B is 25.5 gand its initial temperature is...

Review I Consta A copper cub0 measuring 1.55 cm on edge and an aluminum cube measuring...

Review I Consta A copper cub0 measuring 1.55 cm on edge and an aluminum cube measuring 1.66 cm on edge are both heated to 60.0 C and submerged in 100.0 ml of water at 21.6 C Part A What is the final temperature of the water when equilibrium is reached? (Assume a density of 0.998 g/mlL for water.) ΑΣφ T C Submit Request Answer Part A A block of copper of unknown mass has an initial temperature of 66.4 C....
A 6.66 g lead weight, initially at 9.3 oC, is submerged in 10.00 g of water...

A 6.66 g lead weight, initially at 9.3 oC, is submerged in 10.00 g of water at 52.3 oC in an insulated container. What is the final temperature of both substances at thermal equilibrium? Specific heat capacity of lead = 0.128 J/gC; water = 4.18 J/gC O A. 52.3 oC OB. 51.4 OC OC. 53.6 oC OD. 49.4 oC OE. 50.3 oC OF. 47.6 OC O G. 48.4 oC OH. None of the above
A hot lump of 32.3 g of copper at an initial temperature of 96.5°C is placed...

A hot lump of 32.3 g of copper at an initial temperature of 96.5°C is placed in 50.0 mL H2O initially at 25.0°C and allowed to reach thermal equilibrium. What is the final temperature of the copper and water given that the specific heat of copper is 0.385J/g°C and the specific heat of water is 4.184J/g°C? 4. A hot lump of 32.3 g of copper at an initial temperature of 96.5°C is placed in 50.0 mL H20 initially at 25.0°C...

A very hot cube of copper metal (32.5 g) is submerged into 105.3 g of water...

A very hot cube of copper metal (32.5 g) is submerged into 105.3 g of water at 15.4 0C and it reach a thermal equilibrium of 17.3 0C. Calculate the initial temperature of copper (Cs = 0.385 J/g 0C for copper, and 4.18 J/g 0C for water)