Question

The specific heat capacity of water is 4.186 J/g °C

Calculate the temperature change when 8000 J of heat is added to 3g of water.
Use the equation q=mcT

Answer 1

Q

C= M×∆T

Step-by-step explanation:

Energy (Q) J

Change of temperature (ΔT) °C

Mass (m) kg

Substance (optional) Custom ▾

Specific heat capacity (c) J/(kg·K)

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Answer 2

Solution given:

specific heat capacity of water [c]4.186 J/g °C

temperature[∆T]=?

mass[m]=3g

heat[Q]=8000J

we have

Q=mc∆T

8000=3*4.186*∆T

∆T=8000/12.558

∆T=637.04°C

the temperature change is 637.04°C.