1)
Heat lost by the copper block = heat gained by water + heat gained by calorimeter
Let the equilibrium temperature be T
"0.024\\times400\\times(210-T)=0.06\\times400\\times(T-31)+0.054\\times4200\\times(T-31)"
"2016-9.6T=24T-744+226.8T-7030.8"
"260.4T=5758.8\n\\\\T=22.12^oC"
2)
"210mm .... 100^oC\\\\\n49mm .... (\\theta)\\\\\n\n0mm .... 0^oC\\\\\n\n\\frac{49 - 0}{210 - 0} =\\frac{\\theta - 0}{100 - 0}\n\n\\\\\\frac{49}{210} = \\frac{\\theta}{100}\n\\\\\n\\theta = 23.3oC"
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