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What is the final tempurature when a 3.0 kg gold bar at 99C is dropped into 0.22 kg of water at 25C Specific heat of gold is 129 j/kg x C

  • physics -

    The sum of the heats gained is zero (something loses heat).

    Heatgainedwater+heatgainedgold=0
    .22*Cw*(Tf-25)+3*Cg*(Tf-99)=0
    solve for Tf

  • physics -

    A*t =v-u at=v-u add u to both side at*u=v-at

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