Question

Calculate the energy required to heat 1.30kg of water from 22.4°C to 34.2°C . Assume the specific heat capacity of water under these conditions is 4.18·J·g−1K−1 . Round your answer to 3 significant digits.

Answer 1

Answer:

The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J

Explanation:

Calorimetry is the measurement of the amount of heat that a body gives up or absorbs in the course of a physical or chemical process.

The sensible heat of a body is the amount of heat received or transferred by a body when undergoing a temperature variation (Δt) without there being a change in physical state. That is, when a system absorbs (or gives up) a certain amount of heat, it may happen that it experiences a change in its temperature, involving sensible heat. Then, the equation for calculating heat exchanges is:

Q = c * m * ΔT

Where Q is the heat or quantity of energy exchanged by a body of mass m, constituted by a substance of specific heat c and where ΔT is the variation in temperature (ΔT=Tfinal - Tinitial).

In this case:

• $c=4.18 \frac{J}{g*K}$

• m= 1.30 kg= 1,300 g (1 kg=1,000 g)
• ΔT= 34.2 °C - 22.4 °C= 11.8 °C= 11.8 °K  Being a temperature difference, it is independent if they are degrees Celsius or degrees Kelvin. That is, the temperature difference is the same in degrees Celsius or degrees Kelvin.

Replacing:

$Q=4.18 \frac{J}{g*K}*1,300 g*11.8 K$

Q= 64,121.2 J

The energy required to heat 1.30 kg of water from 22.4°C to 34.2°C is 64,121.2 J