A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75° C. T is given by: (Given: room temperature = 30°C, specific heat of copper = 0.1 cal g-1 ° C-1)

Question

A copper ball of mass 100 gm is at a temperature T. It is dropped in a copper calorimeter of mass 100 gm, filled with 170 gm of water at room temperature. Subsequently, the temperature of the system is found to be 75° C. T is given by: (Given: room temperature = 30°C, specific heat of copper = 0.1 cal g-1 ° C-1) (A) 800°C (B) 885°C (C) 1250°C (D) 825°C

Tardigrade Admin · Accepted Answer

100 × 0.1 × (t - 75) = 100 × 0.1 × 45 + 170 × 1 × 45 10t - 750 = 450 + 7650 10t = 1200 + 7650 10t = 8850 t = 885° C

$80 0^{\circ} C$

$88 5^{\circ} C$

$125 0^{\circ} C$

$82 5^{\circ} C$

$100 \times 0.1 \times (t - 75) = 100 \times 0.1 \times 45 + 170 \times 1 \times 45$
$10 t - 750 = 450 + 7650$
$10 t = 1200 + 7650$
$10 t = 8850$
$t = 88 5^{\circ} C$

800∘C

885∘C

1250∘C

825∘C

100×0.1×(t−75)=100×0.1×45+170×1×45 10t−750=450+7650 10t=1200+7650 10t=8850 t=885∘C

$80 0^{\circ} C$

$88 5^{\circ} C$

$125 0^{\circ} C$

$82 5^{\circ} C$

$100 \times 0.1 \times (t - 75) = 100 \times 0.1 \times 45 + 170 \times 1 \times 45$
$10 t - 750 = 450 + 7650$
$10 t = 1200 + 7650$
$10 t = 8850$
$t = 88 5^{\circ} C$