Question

The mass specific heat capacity and initial temperature of the sphere was 1000 gm,0.5cal/gm°C and 80°C. The mass of the liquid and calorimeter are 900gm and 200gm, initially both were at room temperature 20°C .Both calorimeter and sphere are made of same material. If the steady state temperature after mixing is found to be 40°C then the specific heat capacity of unknown liquid is?
(1) 0.25 cal/gm°C
(2)0.5 cal/gm°C
(3)1 cal/gm°C
(4)1.5 cal/gm°C​

Answer 1

Answer:

Specific heat capacity of unknown liquid is 1 Cal/g C

Explanation:

Heat given by the sphere is equal to the heat absorbed by the liquid and calorimeter

So we will have

$m_1s_1 \Delta T_1 = m_2 s_2\Delta T_2 + m_3s_1\Delta T_2$

$1000(0.5)(80 - 40) = 900 s (40 - 20) + 200(0.5)(40- 20)$

$20000 = 18000 s + 2000$

so we have

$18000 = 18000 s$

$s = 1 Cal/ g ^oC$

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Topic : Calorimetery

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