The temperature of 0.5 kg of water is 303 K. It is cooled to 278K by keeping it in a refrigerator.what is the time taken by water to reach 278 K if 87.5J of heat is given out in each second? ( specific heat capacity of water is 4200 J/kg K )
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We have : Q = mcθ.
2. θ = (θ2 - θ1) = (278 - 303) = -25 K
• The negative sign indicates that, heat energy is given out
3. Substituting the known values we get: Q = 0.5×4200×-25 = -52500 J
4. Energy given out per second = -87.5 J
5. So time taken = -52500⁄-87.5 = 600 seconds = 10 minutes
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