Question

How much heat is liberated (in kJ) from 249 g of silver when it cools from 87 °C to 26 °C? The heat capacity of silver is 0.235 Jg^{-1} °C^{-1}. Note, "heat liberated" implies that the change in heat is negative. Enter a positive number.

Answer 1

The heat liberated from 249 g of silver when it cools from 87 °C to 26 °C is approximately 3.531435 kJ.

Step-by-step explanation:

To calculate the heat liberated from a substance, we can use the formula:

Q = m * C * ΔT

where:

• Q is the heat liberated
• m is the mass of the substance
• C is the heat capacity of the substance
• ΔT is the change in temperature

Given:

• Mass of silver (m) = 249 g
• Heat capacity of silver (C) = 0.235 J/g °C
• Change in temperature (ΔT) = (26 °C - 87 °C) = -61 °C

Plugging these values into the formula, we can calculate the heat liberated:

Q = 249 g * 0.235 J/g °C * (-61 °C)

Q = -3531.435 J

Since the question asks for the heat in kJ, we can convert the value:

Q = -3531.435 J * (1 kJ / 1000 J) = -3.531435 kJ

Therefore, the amount of heat liberated from 249 g of silver when it cools from 87 °C to 26 °C is approximately 3.531435 kJ.

Answer 2

q = - 3.569KJ 0r 3.569KJ Liberated heat (signifying the change in heat is negative)

Step-by-step explanation:

liberated heat implies that change in heat is negative , therefore

q = -m c ΔT

where, m = mass of the Silver = 249 g

c = specific heat capacity of Silver = 0.235 Jg^{-1} °C^{-1

ΔT = change in temperature = 87°C- 26 °C= 61°C

q = -m x c x ΔT

= - 249 x 0.235 x 61 = - 3569.415J rounded to -3569J

Changing to KJ becomes= -3569/1000= - 3.569 KJ

q = - 3.569KJ 0r 3.569KJ liberated heat.