A lead block drops its temperature by 5.90 degrees celsius when 427 J of heat are removed from it. what is the mass of the block?(unit=kg)


  1. Mass of the block is 0.557 kg


    The quantity of heat (q)  liberated during any process is equal to the product of the mass of the block (m), specific heat (C) and the change in temperature (ΔT). Specific heat of lead is 130 J/kg °C.

    We have to rearrange the equation to find the mass of the block of lead as,

    \begin{array}{l}\boldsymbol{q}=\boldsymbol{m} \times \boldsymbol{C} \times \boldsymbol{\Delta} \mathbf{T}\\ \\\boldsymbol{m}=\frac{\mathbf{q}}{\boldsymbol{c} \times \mathbf{\Delta} \mathbf{T}}\\ \\\mathbf{m}=\frac{427 \mathbf{J}}{13 \mathbf{0}_{\mathbf{4} *}^{\prime} \mathbf{r} \times \mathbf{s} \mathbf{9} \mathbf{0}^{*} \mathbf{c}}=\mathbf{0} . \mathbf{5} \mathbf{5} \mathbf{7} \mathbf{k} \mathbf{g}\end{array}

    Thus the mass (m) = 0.577 kg


Leave a Comment