Activity solve the problems below to practice what you have learned. Remember to use the formula for a heat transferred: Q=mC(T: - T:).
Part A
The specific heat of lead is 0.129 yg°C. Find the amount of heat transferred to lead when 355 grams are heated from 95°C to 140°C.
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2060.775 J

Explanation:

From the question given above, we obtained the following data:

Specific heat capacity (C) = 0.129 J/g°C

Mass (m) = 355 g

Initial temperature (Tᵢ) = 95 °C

Final temperature (Tբ) = 140 °C

Heat (Q) tranfered =?

We can calculate the amount of heat transferred by using the following formula:

Q = mC(Tբ – Tᵢ)

Q = 355 × 0.129 (140 – 95)

Q = 45.795 × 45

Q = 2060.775 J

Thus, 2060.775 J of energy was transferred.

explanation: it gains negative electrons, therefore becoming negatively charged. as charge cannot be created, nor destroyed, only transferred, this leaves the wool positive. now that they have opposite charges, the two items attract.