Question

The energy given off as heat by 300g of an alloy as it cools through 50^OC raises the

temperature of 300g of water from 30^OC to 40 ^OC. What is the specific heat of the alloy (in

cal/g· ^OC)?

Answer 1

$$m_{alloy}c_{alloy}t_{alloy}=m_{water}c_{water}(t_{water,f}-t_{water,i}),$$ $$c_{alloy}=\dfrac{m_{water}c_{water}(t_{water,f}-t_{water,i})}{m_{alloy}t_{alloy}},$$ c_{alloy}=\dfrac{300\ g\cdot1.0\ \dfrac{cal}{g\cdot\!^{\circ}C}\cdot(40^{\circ}C-30^{\circ}C)}{300\ g\cdot50^{\circ}C}=0.2\ \dfrac{cal}{g\cdot\!^{\circ}C}.