1.29 g piece of lead has a temperature of 26.0 °C. If it loses 1.90 J of heat to its surroundings, what is its new temperature.

Question

3 years ago

14

Specific heat of lead is 0.128 J/g°C

1 answer:

malfutka [58] · Answer 1 · 2022-09-28T07:43:31+03:00

The new temperature of the piece of lead will be 37.5°C

SPECIFIC HEAT CAPACITY:

The amount of heat absorbed or released by a substance can be calculated by using the following formula:

Q = m × c × ∆T

Where;

Q = quantity of heat absorbed or released (J)

m = mass of substance (g)

c = specific heat capacity

∆T = change in temperature (°C)

According to this question,

Q = 1.90J

m = 1.29g

T1 = 26°C

T2 = ?

c = 0.128 J/g°C

1.90 = 1.29 × 0.128 × (T2- 26°C)

1.90 = 0.165 (T2 - 26°C)

1.90 = 0.165T2 - 4.29

1.90 + 4.29 = 0.165T2

6.19 = 0.165T2

T2 = 37.5°C

Therefore, the new temperature of the piece of lead will be 37.5°C.