Question

A 26.4 g silver ring (cp = 234 J/kg·°C) is heated to a temperature of 66.2°C and then placed in a calorimeter containing 4.94 ✕ 10-2 kg of water at 24.0°C. The calorimeter is not perfectly insulated, however, and 0.136 J of energy is transferred to the surroundings before a final temperature is reached. What is the final temperature?

Answer 1

Answer:

The final temperature of the mixture = 64.834 °C.

Explanation:

Heat lost by the silver ring = heat gained by the water + heat transferred to the surrounding.

c₁m₁(t₁-t₃) = c₂m₂(t₃-t₂) + Q..............Equation 1

Where c₁ = specific heat capacity of the silver copper, m₁ = mass of the silver copper, t₁ = initial temperature of the silver copper, t₃ = final temperature of the mixture. c₂ = specific heat capacity of water, t₂ = initial temperature of water, m₂ = mass of water, Q = energy transferred to the surrounding.

making t₃ the subject of the equation,

t₃ = [c₁m₁t₁+c₂m₂t₂-Q]/(c₁m₁+c₂m₂)........................ Equation 2

Given: c₁ = 234 J/kg.°C, m₁ = 26.4 g, t₁ = 66.2 °C, c₂ = 4200 J/K.°C, m₂ = 4.92×10⁻² kg, t₂ = 24.0 °C, Q = 0.136 J.

Substituting into equation 2

t₃ = [(234×26.4×66.2)+(4200×0.0492×24)-0.136]/[(234×26.4)+(4200×0.0492)]

t₃ = (408957.12+4959.36-0.136)/(6177.6+206.64)

t₃ = (413916.48-0.136)/6384.24

t₃ = 413916.34/6384.24

Thus the final temperature of the mixture = 64.834 °C.

Answer 2

Based on the calculations, the final temperature is equal to 24.01°C.

Given the following data:

• Mass of ring = 26.4 grams

• Specific heat of silver ring = 234 J/Kg°C.

• Mass of water = $4.94 \times 10^{-2}\; Kg$

• Temperature of silver ring = 66.2°C

• Temperature of water = 24.0°C

• Quantity of energy = 0.136 Joules.

Scientific data:

• Specific heat capacity of water = 4200 J/kg°C

To calculate the final temperature, we would apply the law of conservation of heat energy:

The law of conservation of energy.

The quantity of heat energy lost by the silver ring is equal to the quantity of heat energy gained by water and heat transferred to the surroundings.

Mathematically, this is given by the formula:

$Q_{lost} = Q_{gained}+ Q\\\\M_sC_s(T_s - T_f)= M_wC_w(T_f - T_{w}) + Q$

Substituting the given parameters into the formula, we have;

$0.0264 \times 234 \times (66.2 - T_f)= 4.94 \times 4200 \times (T_f - 24.0) + 0.136\\\\6.1776(66.2 - T_f)= 20748 (T_f - 24.0) + 0.136\\\\408.95712-6.1776T_f = 20748 T_f - 497952+0.136\\\\20748 T_f+ 6.1776T_f = 408.95712 + 497952- 0.136\\\\20754.1776T_f = 498360.82112\\\\T_f=\frac{498360.82112}{20754.1776}$

Final temperature = 24.01°C

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