Specific heat of a water flask

Chemistry Level 2

A flask containing 9.0 × 1 0 2 g 9.0 × 10^2 g of water is heated, and the temperature of the water increases from 21 ° C 21 °C to 81 ° C 81 °C . How much heat did the water absorb?

Assumption: The specific heat of water is equal to 4.184 J g ° C \frac{4.184 J}{g °C}

Answer in k J kJ to 3 significant figures.


The answer is 226.

Kano Boom by Kano Boom , 16, Australia

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1 solution

Tom Engelsman
Feb 3, 2021

We require the specific heat formula:

Q = m c w a t e r Δ T = ( 900 ) ( 4.184 ) ( 81 − 21 ) ⋅ 1 k J 1000 J = 225.936 k J ≈ 226 k J . Q = mc_{water} \Delta T = (900)(4.184)(81-21) \cdot \frac{1 kJ}{1000 J} = 225.936 kJ \approx \boxed{226 kJ}.

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